1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.