Let $X$ and $Y$ be varieties, $\varphi:X\to Y$ a morphism, and $\mathcal{G}$ a sheaf on $Y$. Let $s$ denote the sheafification functor. If $\mathcal{F}$ is a sheaf on $X$, let $\varphi_*\mathcal{F}$ denote the direct image.

We define the inverse image $\varphi^{-1}\mathcal{G}$ by sheafifying the presheaf $\varphi_0^{-1}\mathcal{G}$ given by taking the direct limit of $\mathcal{G}$ on the open subsets of $Y$ containing $\varphi(U)$ for each $U\subseteq X$. This operation is functorial.

My question is, do we have that $$s(\varphi_*\varphi_0^{-1}\mathcal{G})=\varphi_*s(\varphi_0^{-1}\mathcal{G})=\varphi_*\varphi^{-1}\mathcal{G}$$ In other words, does the order in which we do the sheafification matter?

I think that $$s(\varphi_0^{-1}\varphi_*\mathcal{F})=\varphi^{-1}\varphi_*\mathcal{F}$$ since here we are composing the functors in order. I feel like the other case should be true as well, but I've tried using the universal property of sheafification and haven't really got anywhere, I can't find arrows both ways to show an isomorphism.

Any help would be much appreciated.


1 Answer 1


There is always a natural map in one direction: Take the sheafification map $\varphi_{0}^{-1}\mathcal{G} \to s(\varphi_{0}^{-1}\mathcal{G})$ on $X$, push forward to get a map $\varphi_{\ast}\varphi_{0}^{-1}\mathcal{G} \to \varphi_{\ast}s(\varphi_{0}^{-1}\mathcal{G})$ of presheaves on $Y$, then this map factors through a map $s(\varphi_{\ast}\varphi_{0}^{-1}\mathcal{G}) \to \varphi_{\ast}s(\varphi_{0}^{-1}\mathcal{G})$ since $\varphi_{\ast}s(\varphi_{0}^{-1}\mathcal{G})$ is a sheaf.

This map doesn't have to be an isomorphism: Take $X := \operatorname{Spec} k \sqcup \operatorname{Spec} k$ and $Y := \operatorname{Spec} k$, let $\varphi : X \to Y$ be the unique $k$-morphism, and let $\mathcal{G} := \mathbb{Z}$; then $\varphi_{0}^{-1}\mathcal{G}$ is the constant presheaf associated to $\mathbb{Z}$ on $X$, so $\Gamma(Y,s(\varphi_{\ast}\varphi_{0}^{-1}\mathcal{G})) = \Gamma(Y,\varphi_{\ast}\varphi_{0}^{-1}\mathcal{G}) = \mathbb{Z}$ but $\Gamma(Y,\varphi_{\ast}s(\varphi_{0}^{-1}\mathcal{G})) = \Gamma(X,s(\varphi_{0}^{-1}\mathcal{G})) = \mathbb{Z} \times \mathbb{Z}$.

  • $\begingroup$ This is false, since $\Gamma(X,s(\varphi_{0}^{-1}\mathcal{G})) = \mathbb{Z} $. $\endgroup$ Jul 29, 2019 at 7:57
  • 1
    $\begingroup$ Here $s(\varphi_{0}^{-1}\mathcal{G})$ is the constant sheaf on $X$ associated to $\mathbb{Z}$, so its global sections is the direct product of $n$ copies of $\mathbb{Z}$ where $n$ is the number of connected components of $X$. $\endgroup$ Jul 29, 2019 at 8:06
  • $\begingroup$ You are right. I deleted my answer, where my argument was inccorect. $\endgroup$ Jul 29, 2019 at 8:09

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