Pullback of sheaves.

Question

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?

Answer 1

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.

Answer 2

The answer is yes. This is one of the cases where it is better to think as sheaves as étalé spaces.

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: $$ f^*(\mathcal F) =f^*(\mathbb R_Y) = \mathbb R_X $$