This is a doubt I have regarding the question: Example: Push-Forward Sheaf
Let $f: X \to Y$ be a continuous map and $\mathscr{F}$ a sheaf on $X$.
We have that $f_*\mathscr{F}$ is the sheaf on $Y$ defined by $f_*\mathscr{F}(U)=\mathscr{F}(f^{-1}(U))$ and if $U$ and $V \subset U$ are opens of $X$, then $\rho_{UV}=\rho_{f^{-1}(U)f^{-1}(V)}: f_*\mathscr{F}(U)\to f_*\mathscr{F}(V)$.
Example: Let $X= \{P,Q\}$ with the discrete topology and $Y=\{R\}.$ Let $f:X\to Y$ be continuous and let $\underline{\mathbb{R}}$ be the sheaf of locally constant real valued functions on $X$. Then:
$ f_*(\underline{\mathbb{R}})=\left\{\begin{array}{ll} \mathbb{R}^2, & U=\{R\} \\ 0, & U=\emptyset\end{array}\right.$
I have two questions:
- Why do we have that $\underline{\mathbb{R}}(f^{-1}(R)) = \mathbb{R}^2$?
- Could anyone give me another simple example of pushforward sheaf, or perhaps a way to build infinitely other examples?
So here is my understanding about question 1:
We have that $\underline{\mathbb{R}}(f^{-1}(R))= \{g:f^{-1}(R) \to \mathbb{R} \ | \ g \ \textrm{is locally constant}\}=\{g:X \to \mathbb{R} \ | \ g \ \textrm{is locally constant}\}$, since $f^{-1}(R)=\{P,Q\}=X$. Since $P$ and $Q$ are opens of $X$, then $g$ is locally constant if, and only if, $g$ is constant on $P$ and $Q$. For example, the map $\varphi:X \to \mathbb{R},\ p \mapsto 1,\ q \mapsto 2$ is locally constant, so $\varphi \in \underline{\mathbb{R}}(f^{-1}(R))$. How can I conclude that $\underline{\mathbb{R}}(f^{-1}(R)) = \mathbb{R}^2$?