Does sheafication commute with pushforward?

Question

Let $X,Y$ be two topological spaces and $\pi:X\to Y$ a continuous function. Given any abelian group valued presheaf $\mathcal F$ on $X$, are $(\pi_*\mathcal F)^{\text{sh}}$ and $\pi_*(\mathcal F^{\text{sh}})$ isomorphic?

Answer 1

Not in general.

Let $A$ be some nontrivial abelian group, let $Y$ be some nontrivial space, and let $X = \{ p \} \subseteq Y$ be a point in $Y$. Let $\pi : \{ p \} \hookrightarrow Y$ be the inclusion map. Let $\mathcal{F}$ be the constant presheaf on $\{ p \}$ with value $A$. Then:

• $\mathcal{F}^{\mathrm{sh}}$ is the constant sheaf on $\{p\}$ with value $A$, so that $\pi_*(\mathcal{F}^{\mathrm{sh}})$ is the skyscraper sheaf on $Y$ at $p$, defined on open sets by $$\pi_*(\mathcal{F}^{\mathrm{sh}})(U) = \begin{cases} A & \text{if } p \in U \\ 0 & \text{otherwise} \end{cases}$$
• $\pi_*\mathcal{F}$ is the constant presheaf on $Y$ with value $A$, so that $(\pi_*\mathcal{F})^{\mathrm{sh}}$ is the constant sheaf on $Y$ with value $A$.

Answer 2

NO :
Let $X=\mathbb R, Y=\{*\}$ (a singleton set) and let $\pi:X\to Y$ be the unique map.
For $\mathcal F$ take the quotient $\mathcal C/\mathcal C_b$ of the sheaf $\mathcal C$ of continuous functions on $\mathbb R$ by its subpresheaf $\mathcal C_b$ of bounded continuous functions.
Then $\mathcal F^{sh}=0$ so that $\pi_* (\mathcal F^{sh})=0$ whereas $(\pi_* \mathcal F)^{sh}=\pi_* (\mathcal F)=\mathcal C(\mathbb R)/\mathcal C_b(\mathbb R)$ =a huge vector space in which $[f]\neq0 \in \mathcal C(\mathbb R)/\mathcal C_b(\mathbb R)$ for every unbounded continuous function $f\in \mathcal C(\mathbb R)$.