Morphism between sheaves induces "isomorphic"(equivalent here?) maps on stalks

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This is an Exercise 3.7 of Tennison Sheaf Theory. I think I solved the problem correctly but I am confused about the problem statement.

Denote $\Gamma f$ as the induced morphism on sheaf of sections if $f$ is a morphism between sheave spaces $(E,p,X)$ and $(E',p',X)$ where $X$ is the base and $p:E\to X$ and $p':E'\to X$.

Denote $\Gamma E$ as the sheaf of sections over $E$.

"If $f:E\to E'$ be a morphism of sheaves over $X$, then $(\Gamma f)_x:(\Gamma E)_x\to (\Gamma E')_x$ and $f\vert_{p^{-1}(x)}:p^{-1}(x)\to p'^{-1}(x)$ where $x\in X$ are isomorphic maps."

There is no reason to expect stalk maps are isomorphisms by considering $E$ as constant sheaf of $Z$, $E'$ as constant sheaf of $Z_2$, $f:(x,z)\to (x,\bar{z})$ where $\bar{z}\in Z_2$ by standard projection map. I think what he is trying to say those stalks map are equivalent instead. The reason is that there is a natural bijection between $(\Gamma E)_x$ and $p^{-1}(x)$ by sending the section to evaluation of section at that point.