$g: S^1 \rightarrow S^1$ homotopic to the identity map, $i$, must be surjective.
Hello all! I had been working on this problem in preparation for my preliminary exam and had run into some difficulty. I suspected that the issue would arise in their induced homomorphisms yielding different groups. Imagining the case where $g(S^1)$ is simply the circle minus a point, we would have that $g_{*}(\pi_{1}(S^1)) = \{0\}$, yet $i_{*}(\pi_{1}(S^1)) = \mathbb{Z}$.
So I had been trying to prove that, given a path $f: [0,1] \rightarrow S^1$, we have $g_{*}([f]) = [g \circ f] = [i \circ f] = i_{*}([f])$. To construct the path homotopy from $g \circ f$ to $i \circ f = f$, I was going to use map $F': [0,1] \times [0,1] \rightarrow S^1$ defined as $F(f(s),t)$ for some homotopy $F$ of $g$ and $i$. The problem is that there are no restrictions on how $F'$ behaves on the sets $\{0\} \times (0,1)$ and $\{1\} \times (0,1)$ - as far as I can see.
I was wondering if someone can help steer me in the correct direction (which may not even be the one I have chosen).
Thanks in advance!
Suppose that $g:S^1\to S^1$ is not surjective, so that it misses a point $p\in S^1$. Then $g$ has image in $S^1\smallsetminus p$, and it factors through $\mathbb R^1\simeq S^1\smallsetminus p$, which is contractible. It follows $g$ is nullhomotopic, and in particular cannot be homotopic to the identity, because $S^1$ is not contractible.