5 Removed nonsense.

However, in this case (I believe had thought that your claim is this was incorrect. Given a map $f:X\to Y$ of spaces (or schemes or what have you) there is an induced geometric morphism of toposes $\text{Sh}(X)\to\text{Sh}(Y)$. That is, there is are functors $f_{*}:\text{Sh}(X)\to\text{Sh}(Y)$ (direct image) and $f^{*}:\text{Sh}(Y)\to\text{Sh}(X)$ (inverse image) such that $f^{*}$ is a left exact left adjoint to $f_{*}$.

(Concretely, the inverse image is a Kan extension and it can be computed as such.) Note though that, regardless of but I had misunderstood precisely how you want to were claiming one should compute it, $f^{*}$ must, qua left adjoint, preserve colimits.

So, fix two open sets $U$ and $V$ in $Y$. The corresponding representable functors are sheaves, but we can easily choose $Y,U,V$ so that their presheaf sum $U+V$ is not yet a sheaf. I.e., in order to obtain the sheaf sum we must first sheafify: $a(U+V)$. As I have said, the inverse image must preserve colimits and so it must be $a(f^{-1}(U)+f^{-1}(V))$ since this is the sum in $\text{Sh}(X)$ of the open sets $f^{-1}(U)$ and $f^{-1}(V)$. Just as $a(U+V)$ is in general different from the presheaf sum $U+V$, so too these two sums are in general not the same. To take a concrete this case simply let $f$ be the identity map on $X$, then it is clear that you cannot compute the sum of the representable (sheaves) $U$ and $V$ in the way you so have suggested despite the fact that $f$ is an isomorphism of spaces.removed my counterexample since it did not quite address your question.)

4 Fixed some ugly typesetting.

In any case

However, I think that in this case I believe that your claim is not correct as statedincorrect. Consider Given a map $i:X\to f:X\to Y$ of spaces (or schemes or what have you) there is an induced geometric morphism of toposes $\text{Sh}(X)\to\text{Sh}(Y)$. That is, there is are functors $f_{*}:\text{Sh}(X)\to\text{Sh}(Y)$ (direct image) and $f^{*}:\text{Sh}(Y)\to\text{Sh}(X)$ (inverse image) such that $f^{*}$ is a left exact left adjoint to $f_{*}$.

(Concretely, the inverse image is a Kan extension and it can be computed as such.) Note though that, regardless of how you want to compute it, $f^{*}$ must, qua left adjoint, preserve colimits.

So, fix two open sets $U$ and $V$ in $Y$. The corresponding representable functors are sheaves, but we can easily choose $Y,U,V$ so that their presheaf sum $U+V$ is not yet a sheaf. I.e., in order to obtain the sheaf sum we must first sheafify: $a(U+V)$. NowAs I have said, what is the inverse image $i^{*}(a(U+V))$? It is obtained as follows:

• First write $a(U+V)$ as a colimit of representable sheaves (must preserve colimits and so it already is must be $a(f^{-1}(U)+f^{-1}(V))$ since this is the sheaf colimit).
• Then recreate this colimit sum in $\text{Sh}(X)$: $i^{-1}(U)+i^{-1}(V)$.
• Finally, sheafify this in \text{Sh}(X)$of the open sets$\text{Sh}(X)$: f^{-1}(U)$ and $a\bigl(i^{-1}(U)+i^{-1}(V)\bigr)$.
• If I understand your suggestion correctly, you claim that we can avoid step (3) when our map f^{-1}(V)$. Just as$i$is a closed embedding. However, it a(U+V)$ is fairly easy to see that in general different from the presheaf obtained at the end of step (2) will not sum $U+V$, so too these two sums are in general be a sheafnot the same. E.g., To take a concrete case simply let $X=[0,1]$ and f$be the identity map on$Y=\mathbb{R}$with X$, then it is clear that you cannot compute the sum of the representable (sheaves) $U$ and $V$ some open intervals having non-empty intersection with $X$, then it will still not be possible to amalgamate matching data for $i^{-1}(U)+i^{-1}(V)$ for in the same reasons way you have suggested despite the fact that one can't in general already do this for $U+V$ in presheaves on $Y$.f$is an isomorphism of spaces. 3 added 1189 characters in body I guess that you know this though and are asking about the the case where we are not regarding sheaves as \'{e}tale maps. In this case you are right that some sheafification is required (since the construction of the inverse image along$f$involves taking a colimit in$\text{Sh}(X)$). Unfortunately I don't have In any more insight into case, I think that in this case your claim is not correct as stated. Consider a map$i:X\to Y$of spaces and fix two open sets$U$and$V$in$Y$. The corresponding representable functors are sheaves, but we can easily choose$Y,U,V$so that their presheaf sum$U+V$is not yet a sheaf. I.e., in order to obtain the sheaf sum we must first sheafify$a(U+V)$. Now, what is the inverse image$i^{*}(a(U+V))$? It is obtained as follows: • First write$a(U+V)$as a colimit of representable sheaves (it already is since this is the sheaf colimit). • Then recreate this colimit in$\text{Sh}(X)$:$i^{-1}(U)+i^{-1}(V)$. • Finally, sheafify this in$\text{Sh}(X)$:$a\bigl(i^{-1}(U)+i^{-1}(V)\bigr)$. • If I understand your suggestion correctly, you claim that we can avoid step (3) when our map$i$is a closed embedding. However, it is fairly easy to see that the presheaf obtained at the momentend of step (2) will not in general be a sheaf. E.g., take$X=[0,1]$and$Y=\mathbb{R}$with$U$and$V$some open intervals having non-empty intersection with$X$, then it will still not be possible to amalgamate matching data for$i^{-1}(U)+i^{-1}(V)$for the same reasons that one can't in general already do this for$U+V$in presheaves on$Y\$.

2 Fixed a typo.
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