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Let $h:S^1\to X$ a continuous map. If $h_*:\pi_1(S^1)\to \pi_1(X)$ is the trivial induced homomorphism, then $h$ is homotopic to a point.

I'm starting to study fundamental groups and induced maps by fundamental groups, I need help to solve this question, besides that some of you knows any material online or a good basic book that covers induced maps by fundamental groups? I didn't like Lee's book on this topic, very few informations.


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$h$ is a continuous map of spaces, what does it mean for $h$ to be homotopic to a point? You mean $X$ is homotopy equivalent to a point? – user38268 Nov 13 '12 at 22:11
@BenjaLim I meant $h$ is homotopic to a constant function, sorry about my terminology – user42912 Nov 13 '12 at 22:13
@BenjaLim no, $h:S^1\to X$ – user42912 Nov 13 '12 at 22:20
What about Hatcher? – Dejan Govc Nov 14 '12 at 0:26
@DejanGovc I'm thinking about something more concise, but it's an excellent choice. – user42912 Nov 14 '12 at 3:37
up vote 3 down vote accepted

$\pi_1(X)$ consists of homotopy classes of based maps $S^1 \to X$. The group $\pi_1(S^1) \cong \mathbb{Z}$, where the class of identity map $id: S^1 \to S^1$ is the generator of $\pi_1(S^1)$.

Then $h_* : \pi_1(S^1) \to \pi_1(X)$ sends the generator $[id]$ of $\pi_1(S^1)$ to $[h \circ id] = [h] \in \pi_1(X)$. The triviality of $h_*$ is then equivalent to $[h] = [*] \in \pi_1(X)$, where $* : X \to X$ is the constant map at the basepoint. This is precisely the notion that $h$ is (based) nullhomotopic.

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why $[id]$ is the generator of $\pi_1(S^1)$? – user42912 Nov 14 '12 at 3:48
This is part of the theorem that $\pi_1(S^1) \cong \mathbb{Z}$. It turns out the integer associated to a class of maps $[f]$ is just the "degree" of $f$, which counts how many times $f$ loops fully around $S^1$ in a signed way. It could be unclear there aren't "degree $\frac{1}2$" maps whose square is $id$, but in fact there aren't. One way to see how this works is the idea of covering spaces. – Thomas Belulovich Nov 14 '12 at 4:08
Also, it's not even necessary that $[id]$ is the generator of $\pi_1(S^1)$ for the above argument to work. If $h_*$ is trivial, it's definitely the case that $h_*([id]) = 0$, and that's all we needed. – Thomas Belulovich Nov 14 '12 at 4:10
yes, I've already realized we don't need this argument, thank you! – user42912 Nov 14 '12 at 4:11

Let $\alpha:(I,\partial I)\to(S^1,1)$ be an arbitrary loop. The condition that $h_*$ is trivial means that $$h_*[\alpha]=[h\circ\alpha]=[\epsilon],\tag{1}$$ where $[\alpha]$ denotes the homotopy class $(\operatorname{rel}\partial I)$ of $\alpha$ and $\epsilon:(I,\partial I)\to (X,h(1))$ is the constant loop defined by $\epsilon(t)=h(1)\in X$ for all $t\in I$. Taking $\alpha(t)=e^{2\pi it}$ for $t\in I$, $(1)$ means that we have a homotopy $H:I\times I\to X$, such that $H(t,0)=h(\alpha(t))=h(e^{2\pi i t})$ and $H(t,1)=\epsilon(t)=h(1)$ for all $t\in[0,1]$ and furthermore, this homotopy is $(\operatorname{rel}\partial I)$, which means that $H(0,s)=H(1,s)=h(1)$ for all $s\in I$.

Now, notice that the map $q:I\times I\to S^1\times I$ defined by $q(t,s)=(e^{2\pi i t},s)$ is a quotient map (because it is closed and surjective). This enables us to define a homotopy $K:S^1\times I\to X$ by the formula $K(e^{2\pi i t},s):=H(t,s)$, for $t\in [0,1]$. This is well defined because $H(0,s)=H(1,s)=h(1)$ and continuous because $q$ is a quotient map. But $K(e^{2\pi i t},0) = H(t,0) = h(\alpha(t))=h(e^{2\pi i t})$ and $K(e^{2\pi i t},1)=H(t,1)=h(1)$. Thus, $K$ is a homotopy from $h$ to the constant map $e^{2\pi i t}\mapsto h(1)$, which is exactly what we were trying to find.

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