Suppose to have a continuous function $\alpha:X\to Y$ between topological spaces, and suppose to have two sheaves, $\mathcal F$ defined on $X$ and $\mathcal G$ defined on $Y$. It's well known how are defined the direct image $\alpha_*\mathcal F$, also called pushforward, and the inverse image $\alpha^{-1}\mathcal G$, also called pullback. Clearly, any open set in $Y$ containing $\alpha(x)$ must be of the form $\alpha (U)$ for some open $U\subset X$ containing $x$; in fact, $(\alpha^{-1}\mathcal G)_x$ is equal to $\mathcal G_{\alpha (x)}$.
My problem is that I don't understand why, fixed $x\in X$, in general $(\alpha_*\mathcal F)_x$ is not equal to $(\mathcal F)_x$. I understand that the open sets of the form $\alpha ^{-1}(V)$, with $V\subset Y$ open set containing $\alpha (x)$, are not all the open set of $X$ containing $x$, however the equivalence classes in $(\mathcal F)_x$ seem to me to be all covered by $(\alpha_*\mathcal F)_x$. What I mean is that, for any section $f\in \mathcal F(U)$, with $U\subset X$ an open set containing $x$ that is not preimage of an open of $Y$, one could take $V\subset Y$ open, containing $\alpha (x)$, and consider the restriction of $f$ to $W:=\alpha ^{-1} (V)$. So $f_x=(f|_W)_x\in(\alpha_*\mathcal F)_x$. Where is the contradiction \ misunderstanding in my reasoning? Thanks in advance.
You write:
That's not true. There's no reason why sets of the form $\alpha(U)$ should even be open, and $\alpha(x)$ may have many many open neighborhoods in $Y$ which are not images of open neighborhoods of $x$.
You also write:
To do this, we would have to have $\alpha^{-1}(V)\subset U$. Why do you think that such a $V$ must exist?