# Pullback of sheaves topology and sites

Let $$T = Top$$ be the site of topological spaces with the usual open covering and let $$\mathcal{F}$$ be a sheaf on the site $$T$$. Naturally, for a space $$X \in Ob(T)$$, the sheaf $$\mathcal{F}$$ induces a sheaf in topological sense over $$X$$ in the following way: For $$U \subset X$$ open, we assign the sheaf $$\mathcal{F}_X(U) := \mathcal{F}(U)$$ and the induced restriction maps from $$\mathcal{F}$$. Let $$f : Y \to X$$ be a morphism in $$T$$ (thus a continuous map). One can construct a similar sheaf $$\mathcal{F}_Y$$ over $$Y$$ from $$\mathcal{F}$$.

My question: Is $$\mathcal{F}_Y$$ isomorphic to the pullback sheaf $$f^* \mathcal{F}_X$$?

i.e. $$\mathcal{F}_Y \simeq f^* \mathcal{F}_X$$?

I don't believe this is true: Take $$X=\mathrm{pt}$$ and $$Y$$ the two-point space with one open and one closed point. Then consider the sheaf $$\mathcal F=\hom(-,Y)$$ on the site $$T$$. One then has $$\mathcal F_Y(V)=\hom(V,Y)$$ for any open $$V\subset Y$$. Hence $$\mathcal F_Y=\hom(-,Y)$$. On the other hand, $$f^{\ast}(\mathcal F_X)$$ is the constant sheaf with value $$\rvert Y\lvert$$. But these sheaves do not agree on the closed point $$y$$ of $$Y$$, since $$\lvert (f^{\ast}\mathcal F_X)_y\rvert=2$$ but $$\lvert(\mathcal F_Y)_y\rvert=\lvert\hom(Y,Y)\rvert =3$$ holds.