# Inverse/Direct Images and Sheafification

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.

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 $$\Gamma(Y,\varphi_{\ast}s(\varphi_{0}^{-1}\mathcal{G})) = \Gamma(X,s(\varphi_{0}^{-1}\mathcal{G})) = \mathbb{Z} \times \mathbb{Z}$$.
• This is false, since $\Gamma(X,s(\varphi_{0}^{-1}\mathcal{G})) = \mathbb{Z}$. – Parthiv Basu Jul 29 '19 at 7:57
• 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 $X$. – Minseon Shin Jul 29 '19 at 8:06