I am trying to prove the Vietoris-Begle mapping theorem as indicated in Iversen's Cohomology of Sheaves on page 203. The statement is the following:
Let $f: X\rightarrow Y$ be a proper map between locally compact topological spaces with acyclic fibres and let $G$ be the constant sheaf on Y (with values in an abelian group G, by abuse of notation). Then the induced map $f^{*}: H^{\cdot}(Y,G)\rightarrow H^{\cdot}(X,f^{-1}G)$ is an isomorphism.
As opposed to this statement let me propose a stronger statement:
Let $f:X\rightarrow Y$ be a closed morphism between topological spaces, and let for all $y\in Y$ the canonical map $G\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ be an isomorphism. Then the induced map $f^{*}: H^{\cdot}(Y,G)\rightarrow H^{\cdot}(X,G)$ is an isomorphism.
Now I was wondering why Iversen would impose these additional assumptions. Is there something wrong about my approach?
Proof (of the second statement):
The crucial part is step (4), I have however added steps (1) - (3) in order to increase the readability.
(1) Firstly, since $f$ is closed, so we can deduce that for every open set $V$ of $X$ containing $f^{-1}(y)$ there exists an open set $U$ in $X$ containing $y$, such that $f^{-1}(y)\subseteq f^{-1}(U)\subseteq V$, for example if we set $U:=Y\backslash f(X\backslash V)$.
(2) Now the canonical map $\phi_{y}:(f_{*}f^{-1}G)_{y}\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ is an isomorphism:
Picking some $s\in (f_{*}f^{-1}G)_{y}$, we can choose an open neighbourhood $U$ of $y$ and a section $t\in\Gamma(U;f_{*}f^{-1}G)$ such that $t_{y}=s$. The map $\phi$ sends $s$ to the equivalence class of $t$ in $colim_{f^{-1}(y)\subseteq V\subseteq X \text{ open neighbourhood}} \Gamma(V,f^{-1}G)$. Assume it is sent to zero. This means that there is an open set $V\subseteq X$ containing $f^{-1}(y)$ such that $t_{\vert V}=0$, but we can actually choose some open neighbourhood $W\subseteq Y$ of $y$ such that $f^{-1}(W)\subseteq V$ and $t_{\vert W}=0$, so injectivity is shown.
Surjectivity is similar.
(3) Now we can prove that the unit map $G\rightarrow f_{*}f^{-1}G$ is an isomorphism: pick a point $y$ of $Y$ and observe that the composition $G_{y}\rightarrow (f_{*}f^{-1}G)_{y}\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ is an isomorphism by assumption.
(4) This is where my approach is differing substantially from Iversen's, so if something is wrong about my proof, it will most likely be in this part:
Let for any topological space $a_{Z}:Z\rightarrow *$ be defined. The global section functor can be written as $\Gamma(Z,_)=(a_{Z})_{*}$. The unit map $G\rightarrow f_{*}f^{-1}G$ hence induces a natural isomorphism $\Gamma(Y;G)=(a_{Y})_{*}G\rightarrow (a_{Y})_{*}f_{*}f^{-1} G=(a_{X})_{*}f^{-1}G=\Gamma(X,f^{-1}G)$. This natural transformation is the one which induces the map $f^{*}H^{\cdot}(Y;G)= R\Gamma(Y;G)\rightarrow R\Gamma(X;f^{-1}G)$ by the universal property of the derived functor, and again by the universal property, $f^{*}$ is an isomorphism q.e.d.
I also welcome counterexamples.
Thanks Roland! And here is what is wrong about my reasoning: What I have somehow hidden in the last sentence of my proof with the words "by the universal property" is the following:
Let $Sh(Y)$ the category of sheaves on $Y$ and $Ab$ the category of abelian groups. Both the functors $(a_{Y})_{*}$ and $(a_{Y})_{*}f_{*}f^{-1}$ induce functors $K^{+}(Sh(Y))\rightarrow K(Ab)$ on the category of complexes modulo homotopy (bounded below) as usual. Let $P_{Sh(Y)}:K^{+}(Sh(Y))\rightarrow D^{+}(Sh(Y))$ and $P_{Ab}:K(Ab)\rightarrow D(Ab)$ be the localisation functors. $R(a_{Y})_{*}$ comes equipped with a natural transformation $\varepsilon: P_{Ab}(a_{Y})_{*}\Rightarrow R((a_{Y})_{*})P_{Sh(Y)}$ and $R((a_{Y})_{*}f_{*}f^{-1})$ comes equipped with a natural transformation $\zeta: P_{Ab}(a_{Y})_{*}f_{*}f^{-1}\Rightarrow R((a_{Y})_{*}f_{*}f^{-1}) P_{Sh(Y)}$.
If we denote by $\tau$ the natural transformation $(a_{Y})_{*}\Rightarrow(a_{Y})_{*}f_{*}f^{−1}$, the universal property of the right derived functor yields a natural transformation $R\tau: R((a_{Y})_{*})\rightarrow R((a_{Y})_{*}f_{*}f^{−1})$ such that $\zeta\circ(P_{Ab}\tau) =(R\tau P_{Sh(Y)})\circ\varepsilon$. Of course, $R\tau$ is just $f^{*}$.
EDIT: The rest of this post contains mistakes. For a better understanding of the context of the accompanying comments I will leave the post as it is.
So far so good. At this point I have incorrecty assumed that the construction $\tau\mapsto R\tau$ is functorial. If it were so, then the existence of $\tau^{-1}$ would yield that $R(\tau^{-1})=(R\tau)^{-1}$, hence we could deduce my claim. Roland's counterexample therefore proves that the construction $\tau\mapsto R\tau$ is not functorial in general.