Direct image of a constant sheaf over a closed set

Question

Let $f : X \to Y$ be a continuous map between topological spaces and $S \subset X$ a closed set. I consider sheaves of $k$-vector spaces for a field $k$. In general, I know that there is a morphism of sheaves $$k_{f(S)} \to f_* k_S$$ and I wonder in which case this can be an isomorphism. I suspect that it is the case if $f$ is a projection but can't prove it. Any idea to prove cases where it is an isomorphism are welcome. Thanks a lot.

Answer 1

We may as well assume $S=X$ and $f(S)=Y$, so $f$ is surjective and we are just considering the morphism $k_Y\to f_*k_X$. Whether this morphism is an isomorphism is closely related to the connectedness properties of $X$ and $Y$. For instance, if $U\subseteq Y$ is any connected open set, then $k_Y(U)=k$, and so for this morphism to be an isomorphism, $k_X(f^{-1}(U))=f_*k_X(U)$ must be $k$ as well, meaning $f^{-1}(U)$ must be connected. If $Y$ is locally connected then this condition is sufficient as well, since it suffices to check that $k_Y(U)\to f_*k_X(U)$ is an isomorphism on a basis of open sets.

In particular, if $X=Y\times F$ and $f:Y\times F\to Y$ is the projection, this will pretty much never be true for arbitrary closed $S\subseteq X$. In particular, given any $y\in Y$ and any closed $C\subseteq F$, you could take $S=\{y\}\times C$, and then for the map $k_{f(S)}\to f_*f_S$ to be an isomorphism you would need $C$ to be connected. So as long as $Y$ is nonempty and $F$ has a disconnected closed subset, there will be $S$ for which it is not true.