Pullback of sheaves.

Let $f : X \to Y$ be a continuous map of topological spaces.. Let $\mathcal{F} = \underline{\mathbb{R}}$ be a sheaf of locally constant function from Y to $\mathbb{R}$ on Y. Let $f^* \mathcal{F}$ be the pullback of the sheaf on X. Is the pullback of $\mathcal{F}$, the sheaf of locally constant functions on X?

• $X$ and $Y$ are spaces or sheaves ? – Tsemo Aristide Feb 7 '17 at 10:06
• they are topological spaces. sorry i need to edit. – Chirantan Chowdhury Feb 7 '17 at 10:11
• The answer is "yes", despite a post below falsely claiming that the answer is "no". – Georges Elencwajg Feb 10 '17 at 20:40
• @GeorgesElencwajg can you explain why on a separate answer? – Chirantan Chowdhury Feb 10 '17 at 20:42
• OK, I've written a separate answer. – Georges Elencwajg Feb 10 '17 at 21:05

The constant sheaf $\mathcal F=\mathbb R_Y$on $Y$ with stalk $\mathbb R$ corresponds to the étalé space $Y\times \mathbb R_{disc}$, where $\mathbb R_{disc}$ denotes $\mathbb R$ endowed with the discrete topology.
The étalé space corresponding to $f^*(\mathcal F)$ is the fibre product $X\times_Y (Y\times \mathbb R_{disc})$ and this fibre product is homeomorphic to $X\times \mathbb R_{disc}$.
Now, the étalé space $X\times \mathbb R_{disc}$ on $X$ corresponds to the constant sheaf $\mathbb R_X$ and this shows what you wanted to know: $$f^*(\mathcal F) =f^*(\mathbb R_Y) = \mathbb R_X$$
No, suppose that $Y$ is a point. The pullback $f^*{\cal F}$ is a constant sheaf, and the sheaf of locally functions defined on $X$ is not always the constant sheaf.
• I don't understand what you are trying to say. The constant sheaf $\underline{\mathbb{R}}$ is exactly the sheaf of locally constant function $X\rightarrow\mathbb{R}$. – Roland Feb 7 '17 at 11:14