I have two questions about the proof of Theorem VI.34 in these lecture notes. I have typed all that is relevant, below.

Theorem. (Nielsen-Schreier) Any subgroup of a finitely generated free group is free.

Proof. Let $F$ be the free group on $n$ generators. Let $X$ be the wedge of $n$ circles, and let $b$ be the central vertex. Then, $\pi_1(X,b) \cong F$. Let $H$ be any subgroup of $F$. Then, by Theorem VI.32, there is a based covering $p: (\tilde X, \tilde b) \to (X,b)$ such that $p_\ast \pi_1(\tilde X, \tilde b) = H$. By Corollary VI.18, $p_\ast$ is injective, and so $\pi_1(\tilde X,\tilde b) \cong H$. Theorem VI.32 states that $\tilde X$ is a simplicial complex, and that $p$ is a simplicial map. Since $p$ is a local homeomorphism, $\tilde X$ can contain only zero and one-dimensional simplices. Therefore, $\tilde X$ is a graph, and so by Theorem IV.11, its fundamental group is free.

  1. How does $p$ being a local homeomorphism ensure that $\tilde X$ contains only zero and one-dimensional simplices? A rigorous proof would be preferred, but intuition (which can be translated to a proof) is also welcome.

  2. I believe the author meant to write connected graph instead of just graph. We know that the fundamental group of connected graphs is free, I don't think this holds for graphs in general. If I am right, why is this graph connected?

Thank you!


1 Answer 1

  1. This is perhaps somewhat misguidingly put. The claim is true since the dimension of a simplex is a homeomorphism invariant, but this is non-trivial. Instead, I would refer to the proof of Theorem VI.32, where the simplicial structure explicitly constructed on $\tilde{X}$ has faces with at most as many vertices as a face in $X$ has vertices. Therefore, if $X$ has dimension $\le1$, so does $\tilde{X}$.

  2. It is proven as part of Theorem VI.32 that $\tilde{X}$ is path-connected. However, this really does not matter. Fundamental groups only make sense relative to a basepoint and the fundamental group $\pi_1(\tilde{X},\tilde{b})$ only depends on the connected component of $\tilde{X}$ containing $\tilde{b}$, which is a connected graph even a priori. In particular, the fundamental group of any graph relative to any basepoint is free and this statement really has no more content than when you only state it for connected graphs.

  • $\begingroup$ Why is the dimension of a simplex homeomorphism invariant? $\endgroup$ Oct 27, 2021 at 1:42
  • 1
    $\begingroup$ The dimension of a simplicial complex $X$ can be characterized as the largest value of $n$ for which there exists an $x\in X$ such that the local homology group $H_n(X,X\setminus\{x\})$ is non-trivial. This notion is invariant under homeomorphism. $\endgroup$
    – Thorgott
    Oct 27, 2021 at 21:16

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