$\pi_1(Y,y_0)\neq\{1\}$ for some $(Y,y_0)\implies\pi_1(S^1,1)\neq\{1\}$

Question

Hint: Otherwise $1_S$ is nullhomotopic, where $1_S$ is the identity map on $S^1,$ and this implies that $f=f\circ1_S$ is nullhomotopic for every closed path $f$ in $Y$ at $y_0.$

I do not understand the hint.

I assumed it comes from a proof by contradiction, thus we assume $\pi_1(S^1,1)=\{1\}$.

But this does not imply $1_S$ to be nullhomotopic because having the trivial fundamental group does not imply the space $X$ to be contractible (which is equivalent to $1_X$ to be nullhomotopic.)

Then, how is that all that implies $f=f\circ1_S$ is nullhomotopic for every closed path $f$ ?

Thank you

Randall · Answer 1 · 2019-08-27T16:45:09.477

1

The identity map $S^1 \to S^1$ is a loop in $S^1$ based at $1$. So if $\pi_1(S^1)$ is trivial this loop (the identity map) must be null homotopic.

Finally, loops in $Y$ are based maps $S^1 \to Y$, so the second claim follows: a null map post-composed with any map must be null.

If you do not consider "closed paths in $Y$" to be based maps $S^1 \to Y$, but instead as paths $I \to Y$ which agree at the ends, then pass to the quotient, gluing the ends together to get $S^1$.

edited Aug 27 '19 at 16:45

answered Aug 27 '19 at 16:30

Randall

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Why if $π_1(S^1)$ is trivial this loop (the identity map) must be null homotopic? – Verónica Rmz. Aug 28 '19 at 15:42
Because there's only one class when it's trivial: the null class. – Randall Aug 28 '19 at 15:59
I did find the same question as mine https://math.stackexchange.com/questions/1661942/prove-that-the-fundamental-group-of-the-circle-is-non-trivial-if-the-fundamental – Verónica Rmz. Aug 28 '19 at 16:07
In that answer, Why if $π_1(S^1,1)$ is trivial, then $γ∗$ is a trivial homomorphism? – Verónica Rmz. Aug 28 '19 at 16:12
Because $\gamma_*$ is a homomorphism with domain the trivial group! What else could it be? – Randall Aug 28 '19 at 16:16

$\pi_1(Y,y_0)\neq\{1\}$ for some $(Y,y_0)\implies\pi_1(S^1,1)\neq\{1\}$

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