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I recently started reading Algebraic Topology from Part II of Munkres' book Topology(Second Edition).

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).

I was able to step-by-step verify the proof. But I am oblivious the inutuition behind it. How does the triviality of the induced homomorphism of fundamental groups relates to nullhomotopy is not at all clear to me.

If somebody can provide me with the big picture then it would be great.

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.

caffeinemachine
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    $\pi_1(S^1) \cong \mathbb{Z}$ is generated by the identity $S^1 \to S^1$. To say that the induced map $\pi_1(S^1) \to \pi_1(X)$ is trivial is to say that the image of $S^1$ in $X$ is trivial in $\pi_1$, which means precisely that it's nullhomotopic. (I am ignoring basepoints here but they turn out not to matter in this case.) What's the problem? – Qiaochu Yuan Jun 13 '15 at 06:53
  • What I gather from your comment is that "The image under $h$ of a loop in $S^1$ is path-homotopically the constant loop." This is clear to me. I do not see as clearly how this implies that the map $h$ is null-homotopic. – caffeinemachine Jun 13 '15 at 07:00
  • Consider the identity loop of $S^1$. – archipelago Jun 13 '15 at 08:47
  • @archipelago I had this question from Qiaochu's comment too. What is meant by the identity loop? Is it the function $id:S^1\to S^1$ which carries $x$ to $x$ for all $x$. If yes, then my question is that how is this a loop. A loop is a path with the same starting and ending points. A path has $I=[0, 1]$ as it's domain. The domain of $id:S^1\to S^1$ is not $I$. Can you please address this? – caffeinemachine Jun 13 '15 at 13:14
  • Consider $[0,1]\rightarrow S^1, t\mapsto e^{t2\pi i}$. – archipelago Jun 13 '15 at 14:54
  • @caffeinemachine: a loop based at a point $x \in X$ is the same thing as a map $S^1 \to X$ sending some fixed basepoint of $S^1$ to $x$. – Qiaochu Yuan Jun 14 '15 at 05:23

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I think the origin of the confusion in the comments is that Munkres denotes loops as maps from intervals, instead of maps from $S^1$. A map $\sigma : [0, 1] \to X$ with $\sigma(0) = \sigma(1) = x_0$ is, however, equivalent to a map $\gamma : (S^1, b_0) \to (X, x_0)$ coming from the universal property of quotient spaces, where the domain circle is the quotient space $[0, 1]/\!\!\sim$, $\sim$ pasting $0$ and $1$ to the point $b_0$. Nevertheless, I'll use the notation you're comfortable with :

$\gamma : [0, 1] \to S^1$ be any loop in $S^1$ based at $\gamma(0) = \gamma(1) = b_0$. As $h_*$ is trivial, $h \circ \gamma : [0, 1] \to X$ based at $h(\gamma(0)) = h(\gamma(1)) = x_0$ can be path-homotoped to the zero loop $e_{x_0}$ at $x_0$. Let $F : [0, 1]\times [0, 1] \to X$ denote the path homotopy between $h \circ \gamma$ and $e_{x_0}$.

The desired nullhomotopy then is $H : S^1 \times [0, 1] \to X$ obtained from the universal property of quotient maps, where $S^1 \times [0, 1]$ is considered as the quotient space $[0, 1] \times [0, 1]/\!\! \sim$ under the identification $(0, x) \sim (1, x)$ for all $x$.

Geometrically, this means we are taking the loop in $(X, x_0)$, and letting the intermediates of the nullhomotopy to be the intermediates of the path-homotopy between $\gamma$ and the zero loop. That's all there is to it, as Qiaochu mentioned in the comments.

Balarka Sen
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