Proof. We give the proof for the derived categories, it is the same for spectra.

Since $f_{\ast }$ preserves colimits it affords a right adjoint $f^{!}$ . Then $f_{\ast }\vdash f^{!}$ is a geometric morphism in the opposite direction (note that $f_{\ast }$ preserves finite limits, and in fact all limits, since it is a right adjoint) and consequently $f_{\ast }$ is bi-Quillen. It follows that $f_{\ast }:C^{\text{s}}({\mathcal{D}})\rightarrow C^{\text{s}}({\mathcal{C}})$ preserves weak equivalences, and consequently coincides (up to weak equivalence) with its derived functor.

Now to show that $Lf^{\ast }$ is fully faithful we need to show that $Rf_{\ast }Lf^{\ast }\simeq \operatorname{id}$ . But $Rf_{\ast }\simeq f_{\ast }$ since $f_{\ast }$ is bi-Quillen. Let $E\in C^{\text{s}}({\mathcal{C}})$ be cofibrant. Then $Lf^{\ast }E\simeq f^{\ast }E$ and consequently $Rf_{\ast }Lf^{\ast }E\simeq f_{\ast }f^{\ast }E$ . Since $f^{\ast }$ is fully faithful we have $f_{\ast }f^{\ast }E\cong E$ . This concludes the proof.◻

Proof. For derived categories, this result is classical. For $\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ , the result is also fairly well known, but the author does not know an explicit reference, so we sketch a proof.

Note that there is a Quillen adjunction (in the local model structures)

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}^{\infty }:\operatorname{sPre}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{\ast }\leftrightarrows {\mathcal{S}}{\mathcal{H}}({\mathcal{C}}_{\unicode[STIX]{x1D70F}}):\unicode[STIX]{x1D6FA}^{\infty }.\end{eqnarray}$$

By direct computation using the above adjunction, we find that $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(\unicode[STIX]{x1D6FA}^{\infty }E)=\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)$ , for $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ and $i\geqslant 0$ .

By [Reference LurieLur16, Proposition 1.4.3.4 and Remark 1.4.3.5] the category $\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ admits a $t$ -structure,Footnote ¹ where $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{{\leqslant}0}$ if and only if $\unicode[STIX]{x1D6FA}^{\infty }(E)\simeq \ast$ , and the subcategory $\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{{\geqslant}0}$ is generated under homotopy colimits and extensions by $\unicode[STIX]{x1D6F4}^{\infty }{\mathcal{C}}_{+}$ . We first need to show that this is the $t$ -structure we want, i.e. that the positive and negative parts are determined by vanishing of homotopy sheaves. Since $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(\unicode[STIX]{x1D6FA}^{\infty }E)=\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)$ , this is correct for the negative part. I claim that if $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{{\geqslant}0}$ , then $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)=0$ for $i<0$ . If $X\in \operatorname{sPre}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{\ast }$ , then $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(\unicode[STIX]{x1D6F4}^{\infty }X)=0$ for $i<0$ by direct computation. It thus remains to show that the subcategory of $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ with $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)=0$ for $i<0$ is closed under homotopy colimits and extensions. For extensions this is clear. Homotopy colimits are generated by pushouts and filtered colimits [Reference LurieLur09, Propositions 4.4.2.6 and 4.4.2.7], so we need only deal with cones and filtered colimits. For cones this is again clear, and for filtered colimits it holds because homotopy groups of spectra commute with filtered colimits, and hence the same is true for homotopy sheaves (see the proof of Corollary 3 for more details on this). This proves the claim. Conversely, let $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ with $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)=0$ for $i<0$ . Consider the decomposition $E_{{\geqslant}0}\rightarrow E\rightarrow E_{{<}0}$ . Then $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E_{{\geqslant}0})=0$ for $i<0$ , so $0=\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)=\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E_{{<}0})$ for $i<0$ . It follows that $E_{{<}0}\simeq 0$ and so $E\simeq E_{{\geqslant}0}\in \text{SH}(E)_{{\geqslant}0}$ .

The $t$ -structure is non-degenerate because it is defined in terms of homotopy sheaves, and homotopy sheaves detect weak equivalences by definition.

We have an adjunction

$$\begin{eqnarray}M:\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})\leftrightarrows D({\mathcal{C}}_{\unicode[STIX]{x1D70F}}):U.\end{eqnarray}$$

By construction $U$ is $t$ -exact and thus $M$ is right $t$ -exact. Consider the induced adjunction

$$\begin{eqnarray}M^{\heartsuit }:\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})^{\heartsuit }\leftrightarrows D({\mathcal{C}}_{\unicode[STIX]{x1D70F}})^{\heartsuit }:U.\end{eqnarray}$$

By direct computation using the classical Hurewicz isomorphism (and the above adjunction), $\text{}\underline{\unicode[STIX]{x1D70B}}_{0}(UME)=\text{}\underline{\unicode[STIX]{x1D70B}}_{0}(E)$ if $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})_{{\geqslant}0}$ . It follows that $UM^{\heartsuit }\simeq \operatorname{id}$ . Since $U$ is faithful by definition, from this we deduce that $M^{\heartsuit }U\simeq \operatorname{id}$ as well. Thus $\text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})^{\heartsuit }\simeq D({\mathcal{C}}_{\unicode[STIX]{x1D70F}})^{\heartsuit }\simeq \operatorname{Shv}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ , the latter equivalence being classical. Finally if $X\in {\mathcal{C}}$ and $F\in \operatorname{Shv}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ then $[\unicode[STIX]{x1D6F4}^{\infty }X_{+},F[n]]=[\unicode[STIX]{x1D6F4}^{\infty }X_{+},UF[n]]=H_{\unicode[STIX]{x1D70F}}^{n}(X,F)$ , the first equality by definition and the second by adjunction and the same result in $D({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ .◻

Proof. Let us show that (1) reduces to (2). For $E\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ there is a conditionally convergent spectral sequence

$$\begin{eqnarray}H_{\unicode[STIX]{x1D70F}}^{p}(X,\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}E)\Rightarrow [X,E[p+q]].\end{eqnarray}$$

Under our assumptions on the cohomological dimension of $X$ , it converges strongly to the right-hand side. Under the assumption of commutation of cohomology with filtered colimits, by spectral sequence comparison, it thus suffices to show that for $E_{i}\in \text{SH}({\mathcal{C}}_{\unicode[STIX]{x1D70F}})$ we have $\text{}\underline{\unicode[STIX]{x1D70B}}_{n}(\bigoplus _{i}E_{i})=\bigoplus _{i}\text{}\underline{\unicode[STIX]{x1D70B}}_{n}E_{i}$ .

Now we prove (2). For $E\in {\mathcal{S}}{\mathcal{H}}({\mathcal{C}})$ write $\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(E)(X)=\unicode[STIX]{x1D70B}_{j}(E(X))$ ; this defines a presheaf of abelian groups on ${\mathcal{C}}$ . By definition $\text{}\underline{\unicode[STIX]{x1D70B}}_{j}(E)=a_{\unicode[STIX]{x1D70F}}\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(E)$ . Let $\{E_{i}\}_{i}\in {\mathcal{S}}{\mathcal{H}}({\mathcal{C}})$ . Then $\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(\bigoplus _{i}E_{i})=\bigoplus _{i}\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(E_{i})$ , since homotopy groups of spectra commute with filtered colimits. We may assume that all the $E_{i}$ are cofibrant, so their presheaf direct sum coincides with the derived direct sum. In this case it remains to show that

$$\begin{eqnarray}{a_{\unicode[STIX]{x1D70F}}\bigoplus }_{i}\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(E_{i})\cong \bigoplus _{i}a_{\unicode[STIX]{x1D70F}}\text{}\underline{\unicode[STIX]{x1D70B}}_{j}^{p}(E_{i}).\end{eqnarray}$$

(Note that here we write $\bigoplus _{i}$ for both direct sums of presheaves and direct sums of sheaves, depending on whether the terms on the right are presheaves or sheaves.) But this holds for any collection of presheaves on any site (both sides satisfy the same universal property).

The proof for $D$ is the same.◻

Proof. Certainly $Rf_{\ast }$ is left $t$ -exact if $Lf^{\ast }$ is $t$ -exact by adjunction, and $(Rf_{\ast })^{\heartsuit }$ is right adjoint to $(Lf^{\ast })^{\heartsuit }$ , so it suffices to prove the claims for $Lf^{\ast }$ .

Since ${\mathcal{D}}$ has enough points, it is then enough to assume that $\operatorname{Shv}({\mathcal{D}})=\operatorname{Set}$ . (Indeed let $p:\operatorname{Set}\rightarrow \operatorname{Shv}({\mathcal{D}})$ be a point; we will have

$$\begin{eqnarray}p^{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(Lf^{\ast }E)=\unicode[STIX]{x1D70B}_{i}(Lp^{\ast }Lf^{\ast }E)=p^{\ast }f^{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{i}E\end{eqnarray}$$

for all $E\in \text{SH}({\mathcal{C}})$ by applying the reduced case to $p$ and $fp$ which are points of ${\mathcal{D}}$ and ${\mathcal{C}}$ , respectively. Since ${\mathcal{D}}$ has enough points it follows that $\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(Lf^{\ast }E)=f^{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E)$ , as was to be shown.)

Let $p^{\ast }:\operatorname{Shv}({\mathcal{C}})\leftrightarrows \operatorname{Set}:p_{\ast }$ be a point of ${\mathcal{C}}$ . Then $p^{\ast }$ corresponds to a pro-object in ${\mathcal{C}}$ , which is to say that there is a filtered family $X_{\unicode[STIX]{x1D6FC}}\in {\mathcal{C}}$ such that for $F\in \operatorname{Shv}({\mathcal{C}})$ we have $p^{\ast }(F)=\operatorname{colim}_{\unicode[STIX]{x1D6FC}}F(X_{\unicode[STIX]{x1D6FC}})$ [Reference Gabber and KellyGK15, Proposition 1.4 and Remark 1.5].

It follows that for $E\in {\mathcal{S}}{\mathcal{H}}^{\text{s}}({\mathcal{C}})$ we have

$$\begin{eqnarray}\unicode[STIX]{x1D70B}_{i}(p^{\ast }E)=\unicode[STIX]{x1D70B}_{i}(\operatorname{colim}_{\unicode[STIX]{x1D6FC}}E(X_{\unicode[STIX]{x1D6FC}}))\cong \operatorname{colim}_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D70B}_{i}(E(X_{\unicode[STIX]{x1D6FC}}))=p^{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{i}(E),\end{eqnarray}$$

where the isomorphism in the middle holds because homotopy groups commute with filtered colimits of spectra. In particular $p^{\ast }$ preserves weak equivalences and so $p^{\ast }\simeq Lp^{\ast }$ . Thus the previous equation is precisely what we intended to prove.◻

Proof. The functor $r$ is restriction and $e$ is left Kan extension. Since $e$ preserves covers, $r$ preserves sheaves. Moreover $r$ commutes with taking the associated sheaf, because every cover of $Y\in X_{\text{r}\acute{\text{e}}\text{t}}$ in $Sm(X)$ comes from a cover in $X_{\text{r}\acute{\text{e}}\text{t}}$ (because étale morphisms are stable under composition). It follows that $r$ commutes with colimits. Since $e:X_{\text{r}\acute{\text{e}}\text{t}}\rightarrow Sm(X)_{\text{r}\acute{\text{e}}\text{t}}$ preserves pullbacks (and $X_{\text{r}\acute{\text{e}}\text{t}}$ has pullbacks!), the adjunction is a geometric morphism [Sta17, Tag 00X6]. In order to see that $e$ is fully faithful, i.e. $F\rightarrow reF$ an isomorphism for every $F\in \operatorname{Shv}(X_{\text{r}\acute{\text{e}}\text{t}})$ , we note that for the presheaf adjunction $e^{p}:\operatorname{Pre}(X_{\text{r}\acute{\text{e}}\text{t}})\leftrightarrows \operatorname{Pre}(Sm(k)):r$ we have $re^{p}F=F$ . Indeed this holds for $F$ representable by definition, every sheaf is a colimit of representables, and $e^{p}$ and $f$ both commute with taking colimits. Finally note that for a sheaf $F$ we have $eF=a_{\text{r}\acute{\text{e}}\text{t}}e^{p}F$ and thus $reF=ra_{\text{r}\acute{\text{e}}\text{t}}e^{p}F=a_{\text{r}\acute{\text{e}}\text{t}}re^{p}F=a_{\text{r}\acute{\text{e}}\text{t}}F=F$ , where we have used again that $r$ commutes with taking the associated sheaf.◻

Proof. The functor is fully faithful by Lemmas 5 and 1. It is $t$ -exact by Lemma 4.◻

Proof. The ‘moreover’ part follows from Lemma 4.

Since $f^{\ast }:Y_{\text{r}\acute{\text{e}}\text{t}}\rightarrow X_{\text{r}\acute{\text{e}}\text{t}}$ preserves covers $f_{\ast }:\operatorname{Pre}(X_{\text{r}\acute{\text{e}}\text{t}})\rightarrow \operatorname{Pre}(Y_{\text{r}\acute{\text{e}}\text{t}})$ preserves sheaves and the morphism is continuous. It is a geometric morphism of sites because $f^{\ast }$ preserves pullbacks [Sta17, Tag 00X6].◻

Proof. The first statement is homotopy invariance, see [Reference ScheidererSch94, Example 16.7.2].

For the second statement, we follow closely that proof. Let $f:X\coprod X\rightarrow X\times (\mathbb{A}^{1}\setminus 0)$ be the canonical map. It suffices to show that $R^{n}f_{\ast }F=0$ for $n>0$ and $R^{0}f_{\ast }F=F$ , where we identify $F$ with its pullback to $X\coprod X$ and $X\times (\mathbb{A}^{1}\setminus 0)$ for notational convenience. All of these statements are local on $X$ , so we may assume that $X$ is affine.

Then one may assume that $F$ is constructible (since $\text{r}\acute{\text{e}}\text{t}$ -cohomology commutes with filtered colimits of sheaves, and all sheaves on a spectral space are filtered colimits of constructible sheaves; see again [Reference ScheidererSch94, Example 16.7.2]). Next, writing $X=\operatorname{Spec}(A)$ as the inverse limit of the filtering system $\operatorname{Spec}(A^{\prime })$ , with $A^{\prime }\subset A$ finitely generated over $\mathbb{Z}$ , and using Proposition (A.9) of [Reference ScheidererSch94, Example 16.7.2], we may assume that $X$ is of finite type over $\mathbb{Z}$ .

But $\operatorname{Sper}(\mathbb{Z})=\operatorname{Sper}(\mathbb{Q})=\operatorname{Sper}(\mathbb{R})$ , whence $H_{\text{r}\acute{\text{e}}\text{t}}^{p}(X,F)=H_{\text{r}\acute{\text{e}}\text{t}}^{p}(X\times _{\mathbb{Z}}\mathbb{R},F)$ , so we may assume that $X$ is of finite type over $\mathbb{R}$ .

We may further assume that $F=M_{Z}$ is the constant sheaf on a closed, constructible subset of $X$ (Proposition (A.6) of [Reference ScheidererSch94, Example 16.7.2]).

It is thus enough to prove the analog of our result for an affine semi-algebraic space $X$ over $\mathbb{R}$ and $F=M$ a constant sheaf. But then $H_{\text{r}\acute{\text{e}}\text{t}}^{\ast }(X,M)=H_{\text{sing}}^{\ast }(X(\mathbb{R}),M)$ [Reference DelfsDel91, Theorem II.5.7] and so on, so this is obvious.◻

Proof. We prove the claim for $\text{SH}$ , the proof we give will work just as well for $D$ . We proceed in several steps.

Step 0. If $g$ is étale, then the claim follows from the observation that $f^{\ast }g_{\#}=g_{\#}^{\prime }{f^{\prime }}^{\ast }$ .

Step 1. If $f:X\rightarrow Y$ is any morphism and $E\in \text{SH}(X_{\text{r}\acute{\text{e}}\text{t}})$ , then there is a conditionally convergent spectral sequence

$$\begin{eqnarray}E_{2}^{pq}=R^{p}f_{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}E\Rightarrow \text{}\underline{\unicode[STIX]{x1D70B}}_{-p-q}(Rf_{\ast }E).\end{eqnarray}$$

For this, let $E\in \operatorname{Spt}(X_{\text{r}\acute{\text{e}}\text{t}})$ also denote a fibrant model. Then $Rf_{\ast }E\simeq f_{\ast }E$ and for $U\in Y_{\text{r}\acute{\text{e}}\text{t}}$ we have $f_{\ast }(E)(U)=E(f^{\ast }U)$ . Since $E$ is fibrant there is a conditionally convergent descent spectral sequence

$$\begin{eqnarray}H^{p}(f^{\ast }U,\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}(E))\Rightarrow \unicode[STIX]{x1D70B}_{-p-q}(E(f^{\ast }U)).\end{eqnarray}$$

By varying $U$ , this yields a presheaf of spectral sequences on $Y_{\text{r}\acute{\text{e}}\text{t}}$ . Equivalently, this is a spectral sequence of presheaves. Taking the associated sheaf on both sides we obtain a conditionally convergent spectral sequence

$$\begin{eqnarray}a_{\text{r}\acute{\text{e}}\text{t}}H_{\text{r}\acute{\text{e}}\text{t}}^{p}(f^{\ast }\bullet ,\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}(E))\Rightarrow \text{}\underline{\unicode[STIX]{x1D70B}}_{-p-q}(f_{\ast }E).\end{eqnarray}$$

It remains to see that $a_{\text{r}\acute{\text{e}}\text{t}}H_{\text{r}\acute{\text{e}}\text{t}}^{p}(f^{\ast }\bullet ,F)=R^{p}f_{\ast }F$ , for any sheaf $F$ on $X_{\text{r}\acute{\text{e}}\text{t}}$ . For this we view $F\in D(X_{\text{r}\acute{\text{e}}\text{t}})^{\heartsuit }$ . Then by definition $R^{p}f_{\ast }F=\text{}\underline{\unicode[STIX]{x1D70B}}_{-p}Rf_{\ast }F$ . Repeating the above argument with $D(X_{\text{r}\acute{\text{e}}\text{t}})$ in place of $\text{SH}(X_{\text{r}\acute{\text{e}}\text{t}})$ , we obtain a conditionally convergent spectral sequence

$$\begin{eqnarray}a_{\text{r}\acute{\text{e}}\text{t}}H_{\text{r}\acute{\text{e}}\text{t}}^{p}(f^{\ast }\bullet ,\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}F)\Rightarrow R^{p+q}f_{\ast }F.\end{eqnarray}$$

Since $\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}F=0$ for $q\neq 0$ this spectral sequence converges strongly, yielding the desired identification.

Step 2. If $f$ is proper and of relative dimension at most $n$ , then for $F\in \operatorname{Shv}(X_{\text{r}\acute{\text{e}}\text{t}})$ and $p>n$ we have $R^{p}f_{\ast }F=0$ .

Indeed in this situation, by the proper base change theorem in real étale cohomology [Reference ScheidererSch94, Theorem 16.2], for any real closed point $y\rightarrow Y$ we get $(R^{p}f_{\ast }F)_{y}=H_{\text{r}\acute{\text{e}}\text{t}}^{p}(X_{y},F|_{X_{y}})$ . Since real closed fields are the stalks of the rét-topology, in order for a sheaf $G\in \operatorname{Shv}(Y_{\text{r}\acute{\text{e}}\text{t}})$ to be zero it is necessary and sufficient that $G_{y}=0$ for all such $y$ . But real étale cohomological dimension is bounded by Krull dimension [Reference ScheidererSch94, Theorem 7.6], so we find that $R^{p}f_{\ast }F=0$ for $p>n$ , as claimed.

Conclusion of proof. Since isomorphism in $\text{SH}(Y_{\text{r}\acute{\text{e}}\text{t}}^{\prime })$ is local on $Y^{\prime }$ , it is an easy consequence of step 0 that we may assume that $Y^{\prime }$ is quasi-compact (e.g. affine). Then $f^{\prime }$ is of bounded relative dimension (being of finite type).

Now let $E\in \text{SH}(X_{\text{r}\acute{\text{e}}\text{t}})$ . By $t$ -exactness of $g^{\ast }$ and ${g^{\prime }}^{\ast }$ we get from step 1 conditionally convergent spectral sequences

$$\begin{eqnarray}g^{\ast }R^{p}f_{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}E\Rightarrow \text{}\underline{\unicode[STIX]{x1D70B}}_{-p-q}(g^{\ast }Rf_{\ast }E)\end{eqnarray}$$

and

$$\begin{eqnarray}R^{p}f_{\ast }^{\prime }{g^{\prime }}^{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}E\Rightarrow \text{}\underline{\unicode[STIX]{x1D70B}}_{-p-q}(Rf_{\ast }^{\prime }{g^{\prime }}^{\ast }E).\end{eqnarray}$$

The exchange transformation $g^{\ast }Rf_{\ast }(E)\rightarrow Rf_{\ast }^{\prime }{g^{\prime }}^{\ast }(E)$ induces a morphism of spectral sequences (i.e. respecting the differentials and filtrations). By proper base change for sheaves, we have $g^{\ast }R^{p}f_{\ast }\cong R^{p}f_{\ast }^{\prime }{g^{\prime }}^{\ast }$ . Thus the two spectral sequences are isomorphic. By step 2 the second one converges strongly, and hence so does the first. Thus the result follows from spectral sequence comparison.◻

Remark.

The only place in the above proof where we have used the assumption on $Y$ is in step 1, namely in the construction of the conditionally convergent spectral sequence

$$\begin{eqnarray}R^{p}f_{\ast }\text{}\underline{\unicode[STIX]{x1D70B}}_{-q}E\Rightarrow \text{}\underline{\unicode[STIX]{x1D70B}}_{-p-q}(Rf_{\ast }E).\end{eqnarray}$$

The author does not know how to construct such a spectral sequence in general. He nonetheless contends that the proper base change theorem should be true without assumptions on $Y$ , but perhaps a different proof is needed.

Remark.

In the above proof we deduce proper base change for spectra and unbounded complexes from proper base change for bounded complexes. Since we are dealing with hypercomplete toposes, this is not tautological; see for example [Reference LurieLur09, Counterexample 6.5.4.2 and Remark 6.5.4.3]. The crucial property which seems to make the proof work is encapsulated in step 2 and might be phrased as ‘a proper morphism is locally of finite relative rét-cohomological dimension’. The same is true in étale (instead of real étale) cohomology and this seems to be what the proof of proper base change for unbounded étale complexes [Reference Cisinski and DégliseCD13, Theorem 1.2.1] ultimately rests on, in the guise of [Reference Cisinski and DégliseCD13, Lemma 1.1.7]. This fails for a general proper morphism of topological spaces (consider for example an infinite product of compact positive dimensional spaces mapping to the point).

Proof. Note that $p:W\rightarrow V$ is finite étale, so this makes sense. By continuity (of $F$ ), we may assume that $X$ and $V$ are smooth (and hence so are $Y$ and $W$ ). Write $s:X\rightarrow \operatorname{Spec}(k)$ for the structure map.

If $t:A\rightarrow B$ is any map in $\mathbf{SH}(X)$ , then the canonical diagram

commutes, since $g^{\ast }$ is a functor. Applying this to $\text{tr}_{f}:\unicode[STIX]{x1D7D9}_{X}\rightarrow f_{\#}\unicode[STIX]{x1D7D9}_{Y}$ it is enough to prove that $g^{\ast }(\text{tr}_{f})=\text{tr}_{p}$ under the canonical identifications.

Let $f_{+}:f_{\#}\unicode[STIX]{x1D7D9}_{Y}\simeq \unicode[STIX]{x1D6F4}_{X}^{\infty }Y_{+}\rightarrow \unicode[STIX]{x1D6F4}_{X}^{\infty }X_{+}=\unicode[STIX]{x1D7D9}_{X}$ be the canonical map (so that $\text{tr}_{f}=D(f_{+})$ via $\unicode[STIX]{x1D6FC}_{X,Y}$ ), and similarly for $p_{+}$ . Then $g^{\ast }(f_{+})\simeq p_{+}$ and consequently $g^{\ast }(D(f_{+}))\simeq D(p_{+})$ . It thus remains to show that $\unicode[STIX]{x1D6FC}_{\bullet ,\bullet }$ is natural, i.e. that $g^{\ast }\unicode[STIX]{x1D6FC}_{X,Y}=\unicode[STIX]{x1D6FC}_{V,W}:\unicode[STIX]{x1D6F4}_{V}^{\infty }W_{+}\rightarrow D(\unicode[STIX]{x1D6F4}_{V}^{\infty }W_{+})$ .

For this we use the notation of [Reference Cisinski and DégliseCD12, Example 2.4.3(2), Definition 2.4.24 and Proposition 2.4.31]. The isomorphism $\unicode[STIX]{x1D6FC}_{X,Y}:f_{\#}\unicode[STIX]{x1D7D9}\rightarrow D(f_{\#}\unicode[STIX]{x1D7D9})$ is factored into the isomorphisms $D(f_{\#}\unicode[STIX]{x1D7D9})\rightarrow f_{\ast }\unicode[STIX]{x1D7D9}$ , the Thom transformation $f_{\#}\unicode[STIX]{x1D6FA}_{f}\unicode[STIX]{x1D7D9}\rightarrow f_{\ast }\unicode[STIX]{x1D7D9}$ [Reference Cisinski and DégliseCD12, Definition 2.4.21] and $\unicode[STIX]{x1D6FA}_{f}\unicode[STIX]{x1D7D9}\rightarrow \unicode[STIX]{x1D7D9}$ . All of these are natural in the required sense.◻

Proof. Write $p_{X}:X\times Y\rightarrow X$ and $p_{Y}:X\times Y\rightarrow Y$ for the projections. I claim that the following diagram commutes up to natural isomorphism.

To prove the claim, first note that there is, for $T\in \mathbf{SH}(X),U\in \mathbf{SH}(Y)$ , a natural map $s_{X\times Y\#}(p_{X}^{\ast }T\wedge p_{Y}^{\ast }U)\rightarrow s_{X\#}T\wedge s_{y\#}U$ , which can be obtained by adjunctions, using that the pullback functors are monoidal, and that $s_{X\times Y}=s_{X}\circ p_{X}$ (and similarly for $Y$ ). Then to prove that the comparison map is an isomorphism it suffices to consider $T=\unicode[STIX]{x1D6F4}^{\infty }X^{\prime },U=\unicode[STIX]{x1D6F4}^{\infty }Y^{\prime }$ for $X^{\prime }\rightarrow X$ smooth any $Y^{\prime }\rightarrow Y$ smooth (note that all our functors are left adjoints and so commute with arbitrary sums, and objects of the forms $T,U$ are generators). But then the claim boils down to

$$\begin{eqnarray}X^{\prime }\times _{k}Y^{\prime }\cong (X^{\prime }\times Y)\times _{X\times Y}(X\times Y^{\prime })\end{eqnarray}$$

which is clear.

To prove the lemma, we now specialise to $f:X^{\prime }\rightarrow X$ and $g:Y^{\prime }\rightarrow Y$ finite étale. Then

$$\begin{eqnarray}\text{tr}_{f\times g}=s_{X\times Y\#}(D\unicode[STIX]{x1D6F4}_{X\times Y}^{\infty }(f\times g)_{+}).\end{eqnarray}$$

Note that

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{X\times Y}^{\infty }(f\times g)_{+}=p_{X}^{\ast }\unicode[STIX]{x1D6F4}_{X}^{\infty }f_{+}\wedge p_{Y}^{\ast }\unicode[STIX]{x1D6F4}_{Y}^{\infty }g_{+}.\end{eqnarray}$$

Since $p_{X}^{\ast },p_{Y}^{\ast }$ are monoidal we compute

$$\begin{eqnarray}\text{tr}_{f\times g}=s_{X\times Y\#}p_{X}^{\ast }D\unicode[STIX]{x1D6F4}_{X}^{\infty }f_{+}\wedge p_{Y}^{\ast }D\unicode[STIX]{x1D6F4}_{Y}^{\infty }g_{+}=s_{X\#}D\unicode[STIX]{x1D6F4}_{X}^{\infty }f_{+}\wedge s_{Y\#}D\unicode[STIX]{x1D6F4}_{Y}^{\infty }g_{+},\end{eqnarray}$$

where in the last equality we have used the claim. Since $s_{X\#}D\unicode[STIX]{x1D6F4}_{X}^{\infty }f_{+}=\text{tr}_{f}$ by definition (and similarly for $Y$ ), this is what we wanted to prove.◻

Proof. The usual proof works, see for example [Reference Calmès and FaselCF17, Proof of Corollary 3.5]. We review it. We only show the first statement, the second is similar. Consider the cartesian square

where $\unicode[STIX]{x1D6FF}_{X}:X\rightarrow X\times X$ is the diagonal and similarly for $Y$ . We have the map $\unicode[STIX]{x1D6FD}:\unicode[STIX]{x1D6F4}^{\infty }Y_{+}\wedge \unicode[STIX]{x1D6F4}^{\infty }X_{+}\rightarrow \text{}\underline{K}_{\ast }^{MW}\wedge F\rightarrow F$ , where $\text{}\underline{K}_{\ast }^{MW}\wedge F\rightarrow F$ is the module structure and the first map is the tensor product of $\unicode[STIX]{x1D6F4}^{\infty }Y_{+}\rightarrow \text{}\underline{K}_{\ast }^{MW}$ (corresponding to $a$ ) and $\unicode[STIX]{x1D6F4}^{\infty }X_{+}\rightarrow F$ (corresponding to $b$ ). This defines an element $\unicode[STIX]{x1D6FD}\in F(Y\times X)$ . We have $\text{tr}_{f}((\operatorname{id}\times f)\unicode[STIX]{x1D6FF}_{Y})^{\ast }\unicode[STIX]{x1D6FD}=\text{tr}_{f}(af^{\ast }b)$ and $\unicode[STIX]{x1D6FF}_{X}^{\ast }\text{tr}_{f\times \operatorname{id}_{Y}}\unicode[STIX]{x1D6FD}=\text{tr}_{f}(a)b$ (the latter since $\text{tr}_{\operatorname{id}}=\operatorname{id}$ and $\text{tr}_{f\times g}(x\otimes y)=\text{tr}_{f}(x)\otimes \text{tr}_{g}(y)$ by Lemma 11). These two elements are equal by the base change formula, i.e. Proposition 10.◻

Proof. This is proved for ‘adequate categories of schemes’ in [Reference Cisinski and DégliseCD12, Theorem 2.4.50], of which Noetherian finite dimensional schemes are an example. ◻

Proof. By localisation, we may assume that $X$ is reduced (see for example [Reference Cisinski and DégliseCD12, Proposition 2.3.6(1)]). By [Reference Cisinski and DégliseCD12, Proposition 4.3.9] (this result requires ${\mathcal{M}}$ to come from a model category) we may assume that $X$ is (Henselian) local with closed point $x$ and open complement $U$ . By localisation, it suffices to show that $E|_{x}\simeq 0$ and $E|_{U}\simeq 0$ . The former holds by assumption, and the latter by induction on the dimension. This concludes the proof.◻

Example.

The pseudofunctors $X\mapsto \mathbf{SH}(X)$ and $X\mapsto D_{\mathbb{A}^{1}}(X)$ satisfy the six functors formalism and continuity (for the base category of Noetherian finite dimensional schemes) [Reference AyoubAyo07, Reference Cisinski and DégliseCD12].

Proof. We first show that the images of $G$ in $\operatorname{Ho}({\mathcal{M}}[\unicode[STIX]{x1D6FC}^{-1}])$ are compact homotopy generators. Generation is clear, and for homotopy compactness it is enough to show that a filtered homotopy colimit of $\unicode[STIX]{x1D6FC}$ -local objects is $\unicode[STIX]{x1D6FC}$ -local. But this follows from homotopy compactness of $T\otimes Y^{\otimes n}$ (for $T\in G$ and $n\in \{0,1\}$ ) and definition of $\unicode[STIX]{x1D6FC}$ -locality.

In a model category ${\mathcal{N}}$ with compact homotopy generators, if $T_{1}\rightarrow T_{2}\rightarrow \cdots \,$ is a directed system of weak equivalences then $\operatorname{hocolim}_{i}T_{i}$ is weakly equivalent to $T_{1}$ . (This follows from the same result in the category of simplicial sets.) Thus $U[\unicode[STIX]{x1D6FC}^{-1}]$ is $\unicode[STIX]{x1D6FC}$ -locally weakly equivalent to $U$ .

It remains to see that $U[\unicode[STIX]{x1D6FC}^{-1}]$ is $\unicode[STIX]{x1D6FC}$ -local. This follows from the next two lemmas.◻

Proof. We shall show that (1) $X$ is $\unicode[STIX]{x1D6FC}$ -local if and only if it is $\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC}$ -local, (2) $X$ is $\unicode[STIX]{x1D6FC}^{\prime }$ -local if and only if it is $(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{\prime }$ -local, (3) $X$ is $\unicode[STIX]{x1D6FC}^{\prime }$ -local if it is $\unicode[STIX]{x1D6FC}$ -local and (4) $X$ is $\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC}$ -local if it is $(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{\prime }$ -local.

All tensor products and mapping spaces will be derived in this proof.

(1) Consider the string of maps

$$\begin{eqnarray}\text{Map}(T\otimes Y^{\otimes 3},X)\rightarrow \text{Map}(T\otimes Y^{\otimes 2},X)\rightarrow \text{Map}(T\otimes Y,X)\rightarrow \text{Map}(T,X).\end{eqnarray}$$

If $X$ is $\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC}$ -local, then the composite of any two consecutive maps is an equivalence, and hence all maps are equivalences by 2-out-of-6. Consequently $X$ is $\unicode[STIX]{x1D6FC}$ -local. The converse is clear.

(2) Consider the string of maps

$$\begin{eqnarray}X\rightarrow X\otimes Y\rightarrow X\otimes Y^{\otimes 2}\rightarrow X\otimes Y^{\otimes 3}.\end{eqnarray}$$

If $X$ is $(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{\prime }$ -local then so is $Z\otimes X$ for any $Z$ , since (derived) tensor product preserves weak equivalences. It follows that $X\otimes Y$ is $(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{\prime }$ -local, and hence the composite of any two consecutive maps is an equivalence. Again by 2-out-of-6 this implies that $X$ is $\unicode[STIX]{x1D6FC}^{\prime }$ -local. The converse is clear.

(3) An object $X$ is $\unicode[STIX]{x1D6FC}$ -local if (and only if) for all $T\in {\mathcal{M}}$ the map $\text{Map}(T\otimes Y,X)\rightarrow \text{Map}(T,X)$ is a weak equivalence (of simplicial sets). In particular $T\rightarrow T\otimes Y$ is an $\unicode[STIX]{x1D6FC}$ -local weak equivalence for all $T$ . It also follows that $X\otimes Y$ is $\unicode[STIX]{x1D6FC}$ -local if $X$ is (here we use invertibility of $Y$ ). Since $X\rightarrow X\otimes Y$ is an $\unicode[STIX]{x1D6FC}$ -local weak equivalence, it is a weak equivalence if $X$ (and hence $X\otimes Y$ ) is $\unicode[STIX]{x1D6FC}$ -local. Thus $X$ is $\unicode[STIX]{x1D6FC}^{\prime }$ -local if it is $\unicode[STIX]{x1D6FC}$ -local.

(4) For any simplicial set $K$ we have $[K,\text{Map}(T,X)]=[K\otimes T,X]$ (using a framing if the model category is not simplicial). It follows that $X$ is $\unicode[STIX]{x1D6FC}$ -local if and only if for all $T\in {\mathcal{M}}$ the map $\unicode[STIX]{x1D6FC}^{\ast }:[T\otimes Y,X]\rightarrow [T,X]$ is an isomorphism. In particular, this property can be checked entirely in the homotopy category of ${\mathcal{M}}$ , in which we will work from now on.

Suppose, for now, that $X$ is $\unicode[STIX]{x1D6FC}^{\prime }$ -local. (We will find that our strategy does not work, but it will work for $\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC}$ , and this is all that is left to prove.) We can choose an inverse equivalence $\unicode[STIX]{x1D6FD}:X\otimes Y\rightarrow X$ . We consider the map $\overline{\unicode[STIX]{x1D6FD}}:[T,X]\rightarrow [T\otimes Y,X]$ sending $f:T\rightarrow X$ to $T\otimes Y\xrightarrow[{}]{f\otimes \operatorname{id}}X\otimes Y\xrightarrow[{}]{\unicode[STIX]{x1D6FD}}X$ . We would like to say that $\overline{\unicode[STIX]{x1D6FD}}$ is inverse to $\unicode[STIX]{x1D6FC}^{\ast }$ . Given $f:T\rightarrow X$ we get the following commutative diagram.

Consequently $\unicode[STIX]{x1D6FC}_{\ast }\unicode[STIX]{x1D6FC}^{\ast }\overline{\unicode[STIX]{x1D6FD}}=\unicode[STIX]{x1D6FC}_{\ast }:[T,X]\rightarrow [T,X\otimes Y]$ and thus $\unicode[STIX]{x1D6FC}^{\ast }\overline{\unicode[STIX]{x1D6FD}}=\operatorname{id}$ (note that $\unicode[STIX]{x1D6FC}_{\ast }$ means composition with $X\rightarrow X\otimes Y$ , which is an isomorphism).

The problem is with showing that $\overline{\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D6FC}^{\ast }=\operatorname{id}$ . For this we fix $f:T\otimes Y\rightarrow X$ and consider the following diagram.

If it commutes for all such $f$ , then $\overline{\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D6FC}^{\ast }=\operatorname{id}$ . But this is not clear; the two paths differ by a switch of $Y$ .

However, in any symmetric monoidal category, the switch isomorphism on the square of an invertible object is the identity [Reference DuggerDug14, Propositions 4.20 and 4.21]. Consequently our argument works for $\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC}$ , and this is what we set out to prove.◻

Remark.

The assumption that $Y$ is invertible is necessary in general for the above result. For example, if ${\mathcal{M}}$ is a cartesian symmetric monoidal model category, then there cannot be any $\unicode[STIX]{x1D6FC}^{\prime }$ -local objects unless $\ast =\unicode[STIX]{x1D7D9}\rightarrow Y$ is already an equivalence.

Proof. By the previous lemma, it suffices to show that $X[\unicode[STIX]{x1D6FC}^{-1}]$ is $(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{\prime }$ -local. Clearly $X[\unicode[STIX]{x1D6FC}^{-1}]\simeq X[(\unicode[STIX]{x1D6FC}\otimes \unicode[STIX]{x1D6FC})^{-1}]$ , i.e. we may assume without loss of generality that $Y$ is a square, and so its switch isomorphism (in the homotopy category) is the identity.

Since tensor product commutes with colimits (in each variable) we have $X[1/f]\simeq X\otimes ^{L}\unicode[STIX]{x1D7D9}[1/f]$ , and we can simplify notation by assuming without loss of generality that $X=\unicode[STIX]{x1D7D9}$ .

What we need to prove is that the following diagram induces an equivalence on homotopy colimits.

Because of the domains and codomains, it is tempting to guess that $f_{i}\simeq h_{i}\simeq f_{i}^{\prime }$ . Here we write $f\simeq g$ to mean that the maps become equal in the homotopy category. We claim that this guess is correct. Then if $T$ is any homotopy compact object, applying $[T,\bullet ]$ to our diagram we get a diagram of abelian groups which we need to show induces an isomorphism on colimits. Homotopic maps become equal when applying $[T,\bullet ]$ , and then the desired result follows from an easy diagram chase. By compact generation and stability, this will conclude the proof.

It remains to prove the claim. For this we may work entirely in the homotopy category, which we will do from now on. It is easy to see that indeed $f_{i}=h_{i}$ . For general $Y$ , it would not be true that $f_{i}^{\prime }=f_{i}$ ; one may check that the maps differ by appropriate switches of $Y$ . However, we have assumed that the switch on $Y$ is the identity, so indeed $f_{i}=f_{i}^{\prime }$ as well.◻

Remark.

The stability assumption was used in the above proof in the following form: if $A\rightarrow B$ is any morphism in ${\mathcal{M}}$ and $[T,A]\rightarrow [T,B]$ is an isomorphism for all homotopy compact $T$ , then $A\rightarrow B$ is a weak equivalence. This fails for example in the homotopy category of spaces.

Proof. Recall the element $h\in K_{0}^{MW}(k)$ with the following properties: $\text{}\underline{K}_{n}^{MW}/h=\text{}\underline{I}^{n}$ [Reference MorelMor04, Théorème 2.1] and for $a\in K_{1}^{MW}(k)$ we have $a^{2}h=0$ [Reference MorelMor12, Corollary 3.8] (this relation is the analogue of the fact that in a graded commutative ring $R_{\ast }$ with $a\in R_{1}$ we have $a^{2}=-a^{2}$ by graded commutativity, so $2a^{2}=0$ ). Consequently $\unicode[STIX]{x1D70C}^{2}h=0$ and so $\operatorname{colim}_{n}\text{}\underline{K}_{n}^{MW}\rightarrow \operatorname{colim}_{n}\text{}\underline{I}^{n}$ is an isomorphism. It remains to note that the image of $\unicode[STIX]{x1D70C}$ in $K_{1}^{MW}/h(k)\cong I(k)$ is given by $-(\langle -1\rangle -1)=2\in I(k)\subset W(k)$ , so the induced transition maps in the colimit are precisely those used in Jacobson’s theorem.◻

Proof. It suffices to prove that for a field $k$ , the total signature $W(k)\rightarrow H_{\text{r}\acute{\text{e}}\text{t}}^{0}(k,\mathbb{Z})$ is compatible with transfer. Let $l/k$ be a finite extension and $R/k$ a real closure. There is a commutative diagram

by the base change formula, i.e. Proposition 10. Note that $\operatorname{Sper}(R\otimes _{k}l)$ is the fibre of $\operatorname{Sper}(l)\rightarrow \operatorname{Sper}(k)$ over the ordering corresponding to the inclusion $k\subset R$ . Consequently we also have the following commutative diagram.

Since the signature maps are determined by pulling back to a real closure, this means that we may assume that $k$ is real closed. (Since both sides we are trying to prove equal are additive, we may still assume that $l$ is a field.) But then either $l=k$ or $l=k[\sqrt{-1}]$ . In the former case the transfer on both sides is the identity, and in the latter it is zero.◻

Proof. The map $\coprod _{i}\operatorname{Sper}(l_{i})\rightarrow \operatorname{Sper}(k)$ is a surjective local homeomorphism of compact, Hausdorff, totally disconnected spaces [Reference ScharlauSch85, Theorem 3.5.1 and Remarks 3.5.6]. The result thus follows from the next lemma.◻

Proof. The claim that $\unicode[STIX]{x1D719}$ has finite fibers is well known. We include a proof for convenience of the reader: since $\unicode[STIX]{x1D719}$ is a local homeomorphism the fibers are discrete, since $Y$ is Hausdorff they are closed, and since $X$ is compact they are compact. Now observe that a compact discrete space is finite.

We now prove the surjectivity of the transfer. First we make the following claim: if $X$ is a compact, Hausdorff, totally disconnected space, then given $x\in U\subset X$ with $U$ open, there exists $x\in V\subset U$ such that $V$ is clopen in $X$ . Indeed for $y\neq x$ let $U_{y}$ be a clopen neighbourhood of $y$ disjoint from $x$ . Then $\bigcup _{y\in X\setminus U}U_{y}$ is an open cover of the compact (since closed) complement $X\setminus U$ . Let $U_{1},\ldots ,U_{n}$ be a finite subcover. Then $V=X\setminus \bigcup _{i}U_{i}$ works.

Now consider the morphism $\unicode[STIX]{x1D719}:X\rightarrow Y$ . For $y\in Y$ choose a clopen neighbourhood $U_{y}$ of $y\in Y$ such that there exists a clopen set $V_{y}\subset X$ with $\unicode[STIX]{x1D719}(V_{y})=U_{y}$ and $\unicode[STIX]{x1D719}:V_{y}\rightarrow U_{y}$ a homeomorphism. We will say in this situation that $\unicode[STIX]{x1D719}$ splits strongly over $U_{y}$ . We note that such $V_{y},U_{y}$ exist: since $\unicode[STIX]{x1D719}$ is a local homeomorphism, there exists $V_{y}^{\prime }\subset X$ such that $U_{y}^{\prime }:=\unicode[STIX]{x1D719}(V_{y})$ is an open neighbourhood of $y$ and $\unicode[STIX]{x1D719}:V_{y}^{\prime }\rightarrow U_{y}^{\prime }$ is a homeomorphism. By the claim, we may assume that $V_{y}^{\prime }$ is clopen. Now choose a clopen neighbourhood $U_{y}\subset U_{y}^{\prime }$ , using the claim again. Then $V_{y}:=\unicode[STIX]{x1D719}^{-1}(U_{y})\cap V_{y}^{\prime }$ is clopen in $X$ and maps homeomorphically to $U_{y}$ .

We obtain in this way an open cover $\{U_{y}\}_{y\in Y}$ of $Y$ . Since $Y$ is compact, we can choose a finite subcover $U_{1},\ldots ,U_{n}$ . Using that all the $U_{i}$ are clopen we can refine further until we have found a disjoint clopen cover (replace $U_{i}$ by $U_{i}\setminus (U_{1}\cup U_{2}\cup \cdots \cup U_{i-1})$ ) over which $\unicode[STIX]{x1D719}$ splits strongly. (Note that if $\unicode[STIX]{x1D719}$ splits strongly over a clopen $U\subset Y$ , then it also splits strongly over any clopen $U^{\prime }\subset U$ .)

Since $H^{0}(Y,\mathbb{Z})$ is the set of continuous functions from $Y$ to $\mathbb{Z}$ , it suffices to prove that the indicator function $\unicode[STIX]{x1D712}_{U_{i}}:Y\rightarrow \mathbb{Z}$ of the clopen subset $U_{i}$ is in the image of transfer (because $1=\sum _{i}\unicode[STIX]{x1D712}_{U_{i}}$ , and the transfer is additive). But $\unicode[STIX]{x1D719}$ is strongly split over $U_{i}$ by construction, so there exists some clopen subset $U\subset X$ such that $\unicode[STIX]{x1D719}:U\rightarrow U_{i}$ is a homeomorphism. Then $\unicode[STIX]{x1D712}_{U}\in H^{0}(X,\mathbb{Z})$ and this is taken by transfer to $\unicode[STIX]{x1D712}_{U_{i}}$ , as follows from the explicit description of transfer in terms of ‘summing over preimages’.◻

Proof. The proof can be extracted from the proof of [Sta17, Tag 00VX]. We repeat the argument for convenience. For simplicity, suppose that $U_{\bullet }$ and $V_{\bullet }$ use the same indexing set $I$ , and that the refinement is of the form $V_{i}\rightarrow U_{i}$ . We are given $s_{i}\in F(U_{i})$ for each $i$ , such that $s_{i}|_{U_{i}\times _{X}U_{j}}=s_{j}|_{U_{i}\times _{X}U_{j}}$ , and we need to show that there is a (necessarily unique) $s\in F(X)$ with $s|_{U_{i}}=s_{i}$ .

Let $t_{i}=f^{\ast }s_{i}$ . Then $t_{\bullet }$ is a compatible family for the covering $V_{\bullet }$ , and hence there is $s\in F(X)$ with $s|_{V_{i}}=t_{i}$ for all $i$ . We need to show that also $s|_{U_{i}}=s_{i}$ . For this, fix $i_{0}\in I$ and consider the coverings $U_{\bullet }^{\prime },V_{\bullet }^{\prime }\rightarrow U_{i_{0}}$ obtained by base change. Then $V_{\bullet }^{\prime }$ refines $U_{\bullet }^{\prime }$ . We find that $s_{i_{0}}|_{U_{i_{0}}\times _{X}U_{i}}=s_{i}|_{U_{i_{0}}\times _{X}U_{i}}$ by assumption, and hence $f^{\ast }(s_{i_{0}}|_{U_{i_{0}}\times _{X}U_{i}})=f^{\ast }(s_{i}|_{U_{i_{0}}\times _{X}U_{i}})=t_{i}|_{U_{i_{0}}\times _{X}V_{i}}=s|_{U_{i_{0}}\times _{X}V_{i}}$ by construction. But now because $F$ is separated in the $\unicode[STIX]{x1D70F}$ -topology and $V_{\bullet }^{\prime }\rightarrow U_{i_{0}}$ is a cover we conclude that $s_{i_{0}}=s|_{U_{i_{0}}}$ , as needed.◻

Proof. For this proof, we call a morphism with the properties of $f$ a frét-cover.

The condition is clearly necessary; we show the converse.

$(\ast )$ We first claim that every rét-cover $U_{\bullet }\rightarrow X$ with $X$ smooth Henselian local can be refined by a frét-cover. We can certainly refine $U_{\bullet }$ by an affine cover, so assume that each $U_{i}$ is affine. Then by [Sta17, Tag 04GJ] each $U_{i}$ splits as $U_{i}^{\prime }\coprod U_{i}^{\prime \prime }$ with $U_{i}^{\prime }\rightarrow X$ finite étale and $U_{i}^{\prime \prime }\rightarrow X$ not hitting the closed point $m$ of $X$ (note that $U_{i}\rightarrow X$ is everywhere quasi-finite). I claim that $U_{\bullet }^{\prime }$ is also a rét-cover. Indeed étale morphisms induce open maps on real spectra [Reference ScheidererSch94, Proposition 1.8] and $U_{\bullet }^{\prime }$ covers $R(m)\subset R(X)$ by construction. But the only open subset of $R(X)$ containing $R(m)$ is all of $R(X)$ , by [Reference Andradas, Bröcker and RuizABR12, Propositions II.2.1 and II.2.4]. Finally the real spectrum of any ring is quasi-compact [Reference Andradas, Bröcker and RuizABR12, II.1.5] whence we can always refine by a finite subcover, and then taking the disjoint union we refine by a singleton cover.

If $X\in Sm(k)$ , we write $F_{X}:=F|_{X_{\operatorname{Nis}}}$ for the restriction to the small site. Write $\text{}\underline{\operatorname{Hom}}$ for the internal mapping presheaf functor in this category. Recall that $\text{}\underline{\operatorname{Hom}}(V,F_{X})(V^{\prime })=F(V\times _{X}V^{\prime })$ ; in particular this functor preserves sheaves.

Let $U_{\bullet }\rightarrow X$ be a rét-cover. To show that $F(X)\rightarrow F(U_{\bullet })\rightrightarrows F(U_{\bullet }\times _{X}U_{\bullet })$ is an equaliser diagram (respectively the first map is injective), it is sufficient to show that $F_{X}\rightarrow \text{}\underline{\operatorname{Hom}}(U_{\bullet },F_{X})\rightrightarrows \text{}\underline{\operatorname{Hom}}(U_{\bullet }\times _{X}U_{\bullet },F_{X})$ is an equaliser diagram of sheaves (respectively the first map is an injection of sheaves), since limits of sheaves are computed in presheaves. But finite limits (respectively injectivity) are detected on stalks, whence in both situations we may assume that $X$ is Henselian local. $(\ast \ast )$

Now we show that $F$ is rét-separated. Let $U_{\bullet }\rightarrow X$ be a rét-cover. By the above, to show that $F(X)\rightarrow F(U_{\bullet })$ is injective we may assume that $X$ is Henselian local. Then $U_{\bullet }\rightarrow X$ is refined by a frét-cover $V\rightarrow X$ , by $(\ast )$ . But then $F(X)\rightarrow F(U_{\bullet })\rightarrow F(V)$ is injective since $F$ satisfies the sheaf condition for $V\rightarrow X$ by assumption, so $F(X)\rightarrow F(U_{\bullet })$ is injective and $F$ is rét-separated.

Finally let $U_{\bullet }\rightarrow X$ be any rét-cover. We wish to show that $F$ satisfies the sheaf condition for this cover. By $(\ast \ast )$ , we may assume that $X$ is Henselian local. Then $U_{\bullet }\rightarrow X$ is refined by a frét-cover $V\rightarrow X$ and $F$ satisfies the sheaf condition with respect to $V\rightarrow X$ by assumption, so it satisfies the sheaf condition with respect to $U_{\bullet }\rightarrow X$ by Lemma 23.◻

Remark.

Using [Reference Andradas, Bröcker and RuizABR12, Corollary II.1.15], the claim $(\ast )$ can be extended as follows: Every rét-cover $U_{\bullet }\rightarrow X$ with $X$ arbitrary is refined by a cover $V_{\bullet }^{\prime }\rightarrow V_{\bullet }\rightarrow X$ , where $V_{\bullet }\rightarrow X$ is a Nisnevich cover and each $V_{i}^{\prime }\rightarrow V_{i}$ is a frét-cover.

Proof. We apply Corollary 24. Hence let $\unicode[STIX]{x1D719}:U\rightarrow X$ be a rét-cover with $\unicode[STIX]{x1D719}$ finite étale and $X$ essentially smooth, Henselian local. We need to show that $F$ satisfies the sheaf condition with respect to this cover. Note that $U$ is then a finite disjoint union of essentially smooth, Henselian local schemes, by [Sta17, Tag 04GH (1)].

We now use the transfer $\text{tr}:F_{\ast }(U)\rightarrow F_{\ast }(X)$ from § 4. Any homotopy module is a module over $K_{\ast }^{MW}$ and satisfies the projection formula with respect to this module structure. It follows from Corollary 19 and our assumption that $\unicode[STIX]{x1D70C}$ acts invertibly on $F_{\ast }$ that $F_{\ast }$ is a module over $a_{\text{r}\acute{\text{e}}\text{t}}\mathbb{Z}$ , and satisfies the projection formula with respect to that module structure.

We know that for a Henselian local ring $A$ with residue field $\unicode[STIX]{x1D705}$ , we have $H_{\text{r}\acute{\text{e}}\text{t}}^{0}(A,\mathbb{Z})=H_{\text{r}\acute{\text{e}}\text{t}}^{0}(\unicode[STIX]{x1D705},\mathbb{Z})$ . This follows from [Reference Andradas, Bröcker and RuizABR12, Propositions II.2.2 and II.2.4] (the author learned this argument from [Reference Karoubi, Schlichting and WeibelKSW16, proof of Lemma 6.4]). Consequently by Corollary 21, Proposition 10 and stability of rét-covers under base change, there exists $a\in H_{\text{r}\acute{\text{e}}\text{t}}^{0}(U,\mathbb{Z})$ such that $\text{tr}(a)=1$ .

Now suppose given $b\in F_{\ast }(X)$ such that $b|_{U}=0$ . Then $b=1b=\text{tr}(a)b=\text{tr}(a\cdot b|_{U})=0$ by the projection formula (i.e. Corollary 12). Consequently $F_{\ast }(X)\rightarrow F_{\ast }(U)$ is injective.

Write $p_{1},p_{2}:U\times _{X}U\rightarrow U$ for the two projections and suppose given $b\in F_{\ast }(U)$ such that $p_{1}^{\ast }b=p_{2}^{\ast }b$ . We have to show that there is $c\in F_{\ast }(X)$ such that $b=c|_{U}$ . I claim that $c:=\text{tr}(ab)$ works. Indeed we have $\text{tr}(ab)|_{U}=\unicode[STIX]{x1D719}^{\ast }(\text{tr}_{\unicode[STIX]{x1D719}}(ab))=\text{tr}_{p_{2}}(p_{1}^{\ast }(ab))$ by Proposition 10. Now $p_{1}^{\ast }b=p_{2}^{\ast }b$ by assumption, and so $\text{tr}_{p_{2}}(p_{1}^{\ast }(a)p_{1}^{\ast }(b))=\text{tr}_{p_{2}}(p_{1}^{\ast }(a)p_{2}^{\ast }(b))=\text{tr}_{p_{2}}(p_{1}^{\ast }a)b$ by the projection formula again. Finally $\text{tr}_{p_{2}}(p_{1}^{\ast }a)=\unicode[STIX]{x1D719}^{\ast }\text{tr}_{\unicode[STIX]{x1D719}}(a)=\unicode[STIX]{x1D719}^{\ast }1=1$ by base change again, so we are done.◻

Proof. We follow [Reference HoveyHov01]. Recall that $\operatorname{Spt}({\mathcal{N}},U)$ denotes the category of sequences $(X_{1},X_{2},X_{3},\ldots \,)$ together with bonding maps $X_{i}\otimes U\rightarrow X_{i+1}$ , and morphisms the compatible sequences of morphisms. This is firstly provided with a global model structure $\operatorname{Spt}({\mathcal{N}},U)_{\text{gl}}$ in which a map $(X_{\bullet })\rightarrow (Y_{\bullet })$ is a fibration or weak equivalence if and only if $X_{i}\rightarrow Y_{i}$ is for all $i$ . This is also called a levelwise fibration or weak equivalence. The local model structure is then obtained by localisation at a set of maps which is not important to us, because it only depends on a choice of set of generators of ${\mathcal{M}}$ , and for $L_{H}{\mathcal{M}}$ we can just choose the same generators.

Since in any model category $L_{H_{1}}L_{H_{2}}{\mathcal{N}}=L_{H_{1}\cup H_{2}}{\mathcal{N}}$ , it is enough to show that $L_{H^{\prime }}\operatorname{Spt}({\mathcal{M}},T)_{\text{gl}}=\operatorname{Spt}(L_{H}{\mathcal{M}},T)_{\text{gl}}$ . Note that an acyclic fibration in $\operatorname{Spt}(L_{H}{\mathcal{M}},T)_{\text{gl}}$ is the same as a levelwise acyclic $H$ -local fibration in ${\mathcal{M}}$ , i.e. a levelwise acyclic fibration. Consequently the cofibrations in $\operatorname{Spt}(L_{H}{\mathcal{M}},T)_{\text{gl}}$ are the same as in $\operatorname{Spt}({\mathcal{M}},T)_{\text{gl}}$ , whereas the former has more weak equivalences. Thus the former is a Bousfield localisation of the latter and hence it is enough to show that $L_{H^{\prime }}\operatorname{Spt}({\mathcal{M}},T)_{\text{gl}}$ and $\operatorname{Spt}(L_{H}{\mathcal{M}},T)_{\text{gl}}$ have the same fibrant objects. An object of $\operatorname{Spt}(L_{H}{\mathcal{M}},T)_{\text{gl}}$ is fibrant if and only if it is levelwise $H$ -locally fibrant. An object $E$ of $L_{H^{\prime }}\operatorname{Spt}({\mathcal{M}},T)_{\text{gl}}$ is fibrant if and only if it is levelwise fibrant and $H^{\prime }$ -local, which means that for each $\unicode[STIX]{x1D6FC}:X\rightarrow Y\in H$ and every $n\in \mathbb{Z}$ the map

$$\begin{eqnarray}\text{Map}^{d}(\unicode[STIX]{x1D6F4}^{\infty +n}\unicode[STIX]{x1D6FC},E):\text{Map}^{d}(\unicode[STIX]{x1D6F4}^{\infty +n}Y,E)\rightarrow \text{Map}^{d}(\unicode[STIX]{x1D6F4}^{\infty +n}Y,E)\end{eqnarray}$$

is a weak equivalence. By adjunction, this is the same as $\text{Map}^{d}(\unicode[STIX]{x1D6FC},E_{n})$ being an equivalence, i.e. all $E_{n}$ being $H$ -local. This concludes the proof.◻

Proof. By Lemma 26 we know that $\operatorname{Spt}({\mathcal{S}}{\mathcal{H}}^{S^{1}}(S)[\unicode[STIX]{x1D70C}^{-1}],\mathbb{G}_{m})=\operatorname{Spt}({\mathcal{S}}{\mathcal{H}}^{S^{1}}(S),\mathbb{G}_{m})[\unicode[STIX]{x1D70C}^{-1}]\simeq {\mathcal{S}}{\mathcal{H}}(S)[\unicode[STIX]{x1D70C}^{-1}]$ . But the map $\unicode[STIX]{x1D70C}:S\rightarrow \mathbb{G}_{m}$ is invertible in $\mathbf{SH}^{S^{1}}(S)[\unicode[STIX]{x1D70C}^{-1}]$ and thus $\operatorname{Spt}({\mathcal{S}}{\mathcal{H}}^{S^{1}}(S)[\unicode[STIX]{x1D70C}^{-1}],\mathbb{G}_{m})\simeq {\mathcal{S}}{\mathcal{H}}^{S^{1}}(S)[\unicode[STIX]{x1D70C}^{-1}]$ , i.e. inverting an invertible object has no effect [Reference HoveyHov01, Theorem 5.1].◻

Proof. If $i:Z\rightarrow X$ is a closed immersion then the functor $i_{\ast }:\mathbf{SH}(Z)\rightarrow \mathbf{SH}(X)$ commutes with filtered homotopy colimits (being right adjoint to a functor preserving compact objects) and satisfies $i_{\ast }(X\otimes \mathbb{G}_{m})\simeq i_{\ast }(X)\otimes \mathbb{G}_{m}$ [Reference Cisinski and DégliseCD12, A.5.1(6) and (3)]. It follows from the explicit description of $\unicode[STIX]{x1D70C}$ -localisation in Lemma 15 that $i_{\ast }$ commutes with $L:\mathbf{SH}(X)\rightarrow \mathbf{SH}(X)[\unicode[STIX]{x1D70C}^{-1}]$ . Thus $\mathbf{SH}(X)[\unicode[STIX]{x1D70C}^{-1}]$ satisfies localisation, by [Reference Cisinski and DégliseCD12, Proposition 2.3.19]. Since $\mathbf{SH}(X)[\unicode[STIX]{x1D70C}^{-1}]$ clearly satisfies the homotopy and stability properties, it satisfies the six functors formalism by Theorem 13.

Since $\mathbf{SH}(X)$ is compactly generated so is $\mathbf{SH}(X)[\unicode[STIX]{x1D70C}^{-1}]$ , by the last sentence of Lemma 15.

For any morphism $f:X\rightarrow Y$ the functor $f^{\ast }:\mathbf{SH}(Y)\rightarrow \mathbf{SH}(X)$ commutes with (filtered) homotopy colimits (being a left adjoint), and consequently it commutes with $\unicode[STIX]{x1D70C}$ -localisation, as above. Thus continuity for $\mathbf{SH}(X)[\unicode[STIX]{x1D70C}^{-1}]$ follows from continuity for $\mathbf{SH}(X)$ .◻

Proof. I claim that in $\mathbf{SH}(S)^{\text{r}\acute{\text{e}}\text{t}}$ there is a splitting $\mathbb{G}_{m}\simeq \unicode[STIX]{x1D7D9}\vee \unicode[STIX]{x1D6E5}$ such that the composite $\unicode[STIX]{x1D7D9}\xrightarrow[{}]{\unicode[STIX]{x1D70C}}\mathbb{G}_{m}\simeq \unicode[STIX]{x1D7D9}\vee \unicode[STIX]{x1D6E5}\rightarrow \unicode[STIX]{x1D7D9}$ is the identity. It will follow from Lemma 30 below that $\unicode[STIX]{x1D6E5}\simeq 0$ , proving this lemma.