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