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I'm currently reading the second volume of Voisin's Hodge Theory book and I'm trying to understand some statements in the third chapter, dedicated to monodromy.

In particular, there's the following lemma

Lemma 3.6. Let $\mathcal{F}$ be a local system of stalk $G$ on $X \times [0,1]$. Then $$\mathcal{F} \simeq \operatorname{pr}_1^{-1}(\mathcal{F}|_{X \times 0}).$$

Using this, we would like to prove the main proposition.

Proposition 3.9. If $X$ is connected, locally arcwise connected and simply connected, then every local system $\mathcal{G}$ of stalk $G$ is trivial on $X$, i.e. isomorphic to the constant sheaf $G$.

The proof starts as follows. Fix $x \in X$, and for every $y \in X$, let $\gamma: [0,1] \rightarrow X$ be a path from $x$ to $y$. By lemma 3.6, the inverse image $\gamma^{-1}\mathcal{G}$ is canonically isomorphic to the constant sheaf of stalk $\mathcal{G}_x$, as well as the constant sheaf of stalk $\mathcal{G}_y$.

How is this true? How should I apply lemma 3.6 to get the desired statement?

WindUpBird
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  • If $s\in G$ is viewed as the stalk in $\mathcal{G}_x$, you can use your isomorphism to define the value of $s$ in $y$. The lemma is important to ensure a canonical isomorphism. So to any element of $G$ you can associate a global section of $\mathcal{G}$, which means this local system is trivial – Littlebird Aug 21 '24 at 17:13

1 Answers1

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Your first hypothesis on $X$ ensure the existence of the path $\gamma$, and the last one ensures it is "unique" up to homotopy.

So you need to apply Lemma 3.6 with $X\times [0,1]\times [0,1]$ to prove that everything depends on $\gamma$ only up to homotopy. And as $X$ is simply conected, it doesn't depend on $\gamma$

Construction

Fix a base point $x_0$, and for every $x\in X$ consider a path $\gamma$ joining $x_0$ to $x$.

Let $s\in G$, as $\mathcal{G}$ is a $G$-local system, you can view $s$ as an element of $\mathcal{G}$. As $\mathcal{G}_{|\gamma}$ is trivial (by Lemma 3.6) you can associate to $s$ a canonical element $s_y\in \mathcal{G}_y$.

Let $\mathcal{F}$ be the trivial local system on $X$. I consider the following map of sheaves, defined for connected open subsets (it is enough as $X$ is locally connected) :

To $s\in \mathcal{F}(U)=G$ I can associate elements $s_y\in \mathcal{G}_y$. So I have sections of $\mathcal{G}$ defined on neighborhoods of every points. You can check that this local sections glue together (here you use the independance of $\gamma$), and now you have an element in $\mathcal{G}(U)$.

You can check that this map is an isomorphism on germs.

So you have an isomorphism of sheaves.

Littlebird
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  • I'm sorry if this sound dumb, but I think I don't understand how to use the lemma to conclude exactly that $\mathcal{G}|_\gamma$ is trivial. I mean, I have a sheaf on $X$, what is the sheaf on $X \times [0,1]$? And what does the lemma says about the inverse image $\gamma^{-1}\mathcal{G}$? – WindUpBird Aug 22 '24 at 14:02
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    No problem don't worry. You can apply the lemma to $X$ reduced to a point. Then you have the result on $[0,1]$. That is what I used. Furthermore $\gamma^{-1}\mathcal{G}$ is still a local system. And the lemma say it must be the trivial one. – Littlebird Aug 22 '24 at 15:23