Let $X$ be connected and locally path connected space. Prove that:
$f:X\to S^1$ is null-homotopic iff homomorphism $f_*:\pi_1(X,x)\to\pi_1(S^1,f(x))$ is trivial for some $x$.
I do not have any idea how to prove it.
Prove that $f:X\to S^1$ is null-homotopic iff homomorphism $f_*:\pi_1(X,x)\to\pi_1(S^1,f(x))$ is trivial for some $x$.
algebraic-topologyfundamental-groups
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Best Answer
You say you've seen covering spaces and the lifting lemma but don't have much experience.
To read this answer, first remember what a covering space is.
Then remember the lifting lemma (I don't know whether it's its actual name), which strongly relates covering spaces, the fundamental group and the possibility of lifting maps up a covering map: the lifting property. It says the following thing : if you have a space $S$ (here that's $S^1$), a covering $p:E\to S$ (here $E=\mathbb{R}$ and $p:t\mapsto e^{it}$ will do the trick) and a nice enough space $X$ (nice enough means connected and locally path connected- which are precisely your hypotheses), a point $x_0\in X$, a map $f:X\to S$ and finally a point $e_0\in E$ such that $p(e_0)= f(x_0)$, then you can lift this map up $p$ and send $x_0$ to $e_0$ if and only if $f_\star(\pi_1(X,x_0)) \subset p_\star(\pi_1(E,e_0))$.
Now if $f_\star$ is trivial, then clearly in our situation the inclusion is satisfied, and so we may lift $f$ to $\mathbb{R}$; that is get $g:X\to \mathbb{R}$ with $p\circ g = f$. But then $g$ is null-homotopic, because $\mathbb{R}$ is contractible, so $f$ is null-homotopic as well.
Conversely, if $f$ is null-homotopic, consider $H:X\times [0,1]\to S^1$ a homotopy between $f$ and a constant map. Consider then $i:X \to X\times [0,1]$ defined by $i(x) = (x,0)$. Then $i$ is a deformation retraction and so induces an isomorphism on fundamental groups, hence $H_\star (\pi_1(X\times [0,1], (x_0,0)) = H_\star(i_\star(\pi_1(X,x_0))) = (H\circ i)_\star(\pi_1(X,x_0)) = f_\star(\pi_1(X,x_0))$.
This shows that if $H_\star$ is trivial, then so is $f_\star$. However if we do the same reasoning with $j:X\to X\times [0,1]$ defined by $j(x) = (x,1)$, we get $H_\star(\pi_1(X\times [0,1], (x_0,1)))= 1$ (the trivial group).
Finally we show that $H_\star(\pi_1(X\times [0,1], (x_0,1)))$ and $H_\star (\pi_1(X\times [0,1], (x_0,0))$ are "sort of conjugate". Indeed let $\gamma: [0,1]\to X\times [0,1]$ be a path from $(x_0,0)$ to $(x_0,1)$ (e.g. $\gamma(t) = (x_0,t)$). Then $\pi_1(X\times [0,1], (x_0,1)) = [\gamma^{-1}]\pi_1(X\times [0,1], (x_0,0))[\gamma]$ and so $H_\star(\pi_1(X\times [0,1], (x_0,1)))= H_\star([\gamma^{-1}]\pi_1(X\times [0,1], (x_0,0))[\gamma]) = [H\circ\gamma^{-1}] H_\star(\pi_1(X\times [0,1], (x_0,0)))[H\circ\gamma]$.
This seems very obscure, but then we get that $H_\star(\pi_1(X\times [0,1], (x_0,0)))$ is the trivial group, and so $f_\star(\pi_1(X,x_0))$ as well, and so $f_\star$ is trivial.
(This should be easier to deal with with the notion of a fundamental groupoid instead of the notion of fundamental group: note that the "big problem" for this second part is to manage to deal with the change of basepoints, because a priori the homotopy between $f$ and the constant map does not preserve any base point. I've hinted at this when I said "sort of conjugate". Of course the intuitive idea is the one in Simonsays's answer : since you can contract the image to a single point, you can contract any loop. I think there must be an easier proof than the one I gave here but I'm not sure)