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.