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}$, the sheaf of locally constant functions on X?
Pullback of sheaves.
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algebraic-geometry
sheaf-theory
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1$X$ and $Y$ are spaces or sheaves ? – 2017-02-07
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0they are topological spaces. sorry i need to edit. – 2017-02-07
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0The answer is "yes", despite a post below falsely claiming that the answer is "no". – 2017-02-10
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0@GeorgesElencwajg can you explain why on a separate answer? – 2017-02-10
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0OK, I've written a separate answer. – 2017-02-10
2 Answers
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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: $$ f^*(\mathcal F) =f^*(\mathbb R_Y) = \mathbb R_X $$
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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.
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6I don't understand what you are trying to say. The constant sheaf $\underline{\mathbb{R}}$ is exactly the sheaf of locally constant function $X\rightarrow\mathbb{R}$. – 2017-02-07