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I have a question about the proof of the following result in Kashiwara, Schapira, Sheaves on Manifolds:

Proposition 2.5.12 [Let $X$ be a Hausdorff and locally compact space.] Let $A$ be a ring, and let $M$ be a flat $A$-module. Let $F$ be a sheaf of right $A_X$-modules. Then there is a natural isomorphism: $$ \Gamma_c(X ; F) \underset{A}{\otimes} M \xrightarrow{\cong} \Gamma_c\left(X ; F \underset{A_X}{\otimes} M_X\right). $$

Where given an abelian sheaf $G$ over $X$ we write $\Gamma_c(X,G)$ for the global sections of $G$ with compact support. The text in between brackets is mine and is a permanent assumption as established right before Definition 2.5.5 (p. 104).

The first thing that the proof of the Proposition says is that we “may assume that $X$ is compact,” and then proceeds to prove the result in this case. My question is: Why does it suffice to prove it for $X$ compact? Locally compact Hausdorff spaces are compactly generated, but I don't know how to exploit this here. I also thought that $\Gamma_c(X;F)$ may be written as the filtered colimit of its finitely generated abelian subsheaves, and each such finitely generated abelian subsheaf must have compact support. The thing is that it is not then clear what should we write $\Gamma_c\left(X ; F \underset{A_X}{\otimes} M_X\right)$ a filtered limit of. If $s$ is a global section of $F$, the support of $s\otimes m$ as a section of $F \underset{A_X}{\otimes} M_X$ is contained in $\operatorname{supp}s$. Nonetheless I don't see why the support of every section in $\Gamma_c\left(X ; F \underset{A_X}{\otimes} M_X\right)$ may be written as a finite union of compact sets of the form $\operatorname{supp} s$, where $s\in\Gamma_c(X;F)$. (It might be for finitely many local sections $s$, but not global ones.)

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Any element of $\Gamma_c(X;\mathscr{F})\otimes M$ will, as the finite union - we are a finite sum of pure tensors - of compacta is compact, vanish on restriction to $\Gamma(U;\mathscr{F})\otimes M$ for some $U$ with compact complement. That is a longwinded but precise way of saying: all elements still, meaningfully, can be said to have compact support over some compactum $K\subset X$. Therefore, if an element of $\Gamma_c(X;\mathscr{F})\otimes M$ vanishes in $\Gamma_c(X;\mathscr{F}\otimes\underline{M})$, the corresponding section of $\Gamma(K;\mathscr{F})\otimes M\cong\Gamma_c(K;\mathscr{F}\otimes\underline{M})$ also vanishes and it is not too hard to see that this means the original element was also zero: we consider that restriction onto a partition is an injection, even if we aren't using open subsets, and then use flatness: $$\Gamma_c(X;\mathscr{F})\hookrightarrow\Gamma(X\setminus K;\mathscr{F})\oplus\Gamma(K;\mathscr{F})\\\therefore\Gamma_c(X;\mathscr{F})\otimes M\hookrightarrow(\Gamma(X\setminus K;\mathscr{F})\otimes M)\oplus(\Gamma(K;\mathscr{F})\otimes M)$$and by assumption we would vanish in both components of the direct sum, which contains us, and so we would be zero.

Now, the argument for surjectivity can be made.

Let $\gamma\in\Gamma_c(X;\mathscr{F}\otimes\underline{M})$ be compactly supported over $K$, and thicken $K$ to an open neighbourhood $V$ with $L:=\overline{V}$ being compact, as we may do since we are LCH. We may take $\gamma'\in\Gamma(L;\mathscr{F})\otimes M$ identified with $\gamma|_L$ under the isomorphism of the compact case. It follows that $\gamma'|_{\partial L}=0$, and the extension by zero for closed subspaces gives: $$\Gamma_L(X;\mathscr{F})\supseteq\Gamma_{L\setminus\partial L}(X;\mathscr{F})\cong\ker(\Gamma(L;\mathscr{F})\to\Gamma(\partial L;\mathscr{F}))\\\text{i.e. the following is exact, when $L$ is closed: }\\0\to\Gamma_{L\setminus\partial L}(X;\mathscr{F})\to\Gamma(L;\mathscr{F})\to\Gamma(\partial L;\mathscr{F})$$which is preserved under $(-)\otimes M$ by flatness, and therefore we can find: $$\gamma'\in\Gamma_{L\setminus\partial L}(X;\mathscr{F})\otimes M\subseteq\Gamma_L(X;\mathscr{F})\otimes M\subseteq\Gamma_c(X;\mathscr{F})\otimes M\\\gamma'_L\mapsto\gamma|_L$$ a representative whose support is contained in $L$ and which agrees over $L$ with $\gamma|_L$. The main technical point here is referring to the compactum $\partial L$, because I can know the initial representative $\gamma'$ must vanish on $\partial L$ by naturality and the compact case, but I would have been unable to directly argue - and indeed can't - that $\gamma'|_{L\setminus K}=0$, for example, as $L\setminus K$ is not compact. Even though we demonstrated we always have an injection: $\Gamma_c(U;\mathscr{F})\otimes M\hookrightarrow\Gamma_c(U;\mathscr{F}\otimes M)$, without $U$ being compact, I can't use that here - annoyingly, it is not a given that $\gamma'|_{L\setminus K}$ actually is compactly supported in the appropriate way: we don't know that it lives in $\Gamma_c\otimes M$ rather than $\Gamma\otimes M$.

If $\gamma''$ is the image of $\gamma'$ in $\Gamma_c(X;\mathscr{F}\otimes M)$, we then have that $\gamma''|_L=\gamma|_L$ but, crucially, $\gamma''|_{X\setminus L}=0$ since $\gamma''$ came from a sum of pure tensors of elements of $M$ with elements of $\Gamma_L$, a sum that therefore reduces to the sum $\sum_i 0\otimes m_i=0$ on $X\setminus L$. It follows that $\gamma''=\gamma$ by basic sheaf properties, since $\mathscr{F}\otimes M_X$ is a bona fide sheaf - unlike $\Gamma(-;\mathscr{F})\otimes M$ - so we are done: we found a preimage, and $\Gamma_c(X;\mathscr{F})\otimes M\twoheadrightarrow\Gamma_c(X;\mathscr{F}\otimes M)$ is a surjection and thus an isomorphism.

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  • $\begingroup$ Thanks for the answer! Sorry that it took me a week to review it. I think it is not necessarily true that $\gamma_{i|U_i\cap U_j}=\gamma_{j|U_i\cap U_j}$. Rather, $\gamma_{i|V_k^{ij}}=\gamma_{j|V_k^{ij}}$ for some open cover $U_i\cap U_j=\bigcup_k V_k^{ij}$. This is because the tensor product presheaf is not a separated presheaf in general, so its sheafification must first "separate" the presheaf, see here. See also here for one (rare) example of the tensor product presheaf being separated. $\endgroup$ Commented Nov 25 at 9:59
  • $\begingroup$ @ElíasGuisadoVillalgordo It's a notational error. $\gamma|_{U_i}|_{U_i\cap U_j}=\gamma|_{U_j}|_{U_i\cap U_j}$ holds because $\gamma$ is a section of $\mathscr{F}\otimes M$. However, the representations $\gamma_i\in\Gamma(U_i;\mathscr{F})\otimes M$ may only have that $\gamma_{ij}$ and $\gamma_{ji}$ represent the same section. But it ought to be fixable by compactness of the section and flatness, much as in your linked post. I need to think $\endgroup$ Commented Nov 25 at 10:14
  • $\begingroup$ @ElíasGuisadoVillalgordo I believe I have found a simpler resolution, and improved the rigour of my first paragraph too. There is no sense in rehashing their own proof of the isomorphism when $X$ is compact, which is what I was trying to do. Also, this new simplicity gives me more confidence the authors weren't unfair to implicitly asserted the general case as acceptably similar to the main case. $\endgroup$ Commented Nov 25 at 11:28
  • $\begingroup$ *to have implicitly asserted. Though, now I am struggling to recall their proof of even the compact case - injectivity is easy, surjectivity is not. The obvious attempt runs into the same problem you brought up, and I remember struggling with their proof a year ago or so, and came up with my own "fix" that, as you pointed out, was too brief and was not a fix at all. $\endgroup$ Commented Nov 25 at 11:35
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    $\begingroup$ For a closed $L\subseteq X$ and any space $X$, and any sheaf $\mathscr{F}$ on $X$, we have an exact sequence: $$0\to\Gamma_{\operatorname{int}(L)}(X;\mathscr{F})\to\Gamma(L;\mathscr{F})\to\Gamma(\partial L;\mathscr{F})$$ $\endgroup$ Commented Nov 25 at 14:04

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