Prove that the fundamental group of the circle is non-trivial if the fundamental group of a space $X$ is non-trivial

Question

Prove that if $\pi_1(X,x_0)\neq 1$ for some topological space $X$, then $\pi_1(S^1,1)\neq 1$

I don't quite know how to proceed with this. I know that any path $f:I\to X$ has to factor through $S^1$. I was trying to map non-homotopic paths in $X$, say $f$ and $g$, to $S^1$, and then showing that they're homotopic in $S^1$, and hence must be in $X$. However, I don't know how to map paths from $X$ to $S^1$.

Am I even on the right track? Any help will be greatly appreciated

Pick a non trivial element $S^1 \to X$, and then think about the induced morphism on $\pi_1$ and what happens to the identity. — Justin Young, Feb 18 '16 at 20:08
If the identity $S^1\to S^1$ is nullhomotopic, then so is any map $S^1 \to X$, just compose the homotopy. — Justin Young, Feb 18 '16 at 20:17

score 2 · Answer 1 · answered Feb 18 '16 at 20:26

2

$[\gamma] \in \pi_1(X,x_0) \neq 1$ be a non trivial element. Let $\gamma : S^1 \to X$ be its representation as a loop at $x_0$. Then $\gamma_* : \pi_1(S^1, 1) \to \pi_1(X,x_0)$ be the induced a homomorphism. Now if $\pi_1(S^1,1)$ is trivial then $\gamma_*$ is a trivial homomorphism. But image of $\gamma_*$ contain the element $[\gamma]$. So contradiction.

[As I observe @Justin Young also advice some similar argument in the comment]

answered Feb 18 '16 at 20:26

Anubhav Mukherjee

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Why if $\pi_1(S^1,1)$ is trivial, then $\gamma_*$ is a trivial homomorphism? – Verónica Rmz. Aug 28 '19 at 15:48
what is a homomorphism from a trivial group to some other other group? Is it not the trivial one? – Anubhav Mukherjee Aug 28 '19 at 16:39

Prove that the fundamental group of the circle is non-trivial if the fundamental group of a space $X$ is non-trivial

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