What does the fact that $[p_0]$ generates the fundamental group of $S^1$ mean? I don't know where this fact was used.

A part of Lemma 55.3 in the book proves the following:

Let $h:S^1\to X$ be a continuous function. Assume that $h_*$ is the trivial homomorphism of fundamental groups. Then the map $h:S^1\to X$ is nullhomotopic (The converse is also proven in the lemma).

The lemma is proved as follows in the book:

Let $p:\mathbf R\to S^1$ be the standard covering map $x\mapsto (\cos 2\pi x, \sin 2\pi x)$ and let $p_0=p|_I$, where $I=[0, 1]$. Then $[p_0]$ is a loop in $S^1$ based at $b_0:=(1, 0)$ and it generates $\pi_1(S^1, b_0)$.

Let $x_0=h(b_0)$. Since $h_*$ is trivial, the loop $f:=h\circ p_0$ can be path-homotoped to the constant loop based at $x_0$. So let $F$ be a path homotopy in $X$ between the loops $f$ and the constant loop $e_{x_0}$. Now the map $p_0\times id:I\times I\to S^1\times I$ is a closed continuous sujective map and is thus a quotient map. Also, $(0, t)$ and $(1, t)$ are mapped to $(b_0, t)$ under this map for each $t\in I$. The path homotopy $F$ maps $0\times I$, $1\times I$ and $I\times 1$ to $x_0$ of $X$, and so it induces a continuous map $H:S^1\times I\to X$ that is a homotopy between $h$ and a constant map.


1 Answer 1


I thought about it for a day. it's a guess. I think Munkres said it just to easily prove that the map $H$ is induced by the path homotopy $F$ (Theorem 22.2)

the fact that $[p_0]$ is the generator implies that the map $p_0\times id$ rotates $I\times I$ just once. So $p_0\times id$ has a injectivity except for $0\times t$,$1\times t$.Then ,for each $(s,t)\in S^1\times I$, $s\not= b_0$, $(p_0\times id)^{-1}(s,t)$ is a one point set. Of course, on this one point set, F is constant.

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