This exercise is seemingly very easy:
Let $X$ be a space. Show that: $$\pi_1(X,x_0)=0$$For all $x_0\in X$ iff. all maps $f:S^1\to X$ are nullhomotopic.
The problem: Hatcher remarks at the end of the exercise:
"In this problem, ‘homotopic’ means ‘homotopic without regard to basepoints’"
Which is rather problematic. In $\pi_1(X,x_0)$, the elements are equivalence classes of loops $f:I\to X$, $f(0)=x_0=f(1)$, with equivalence relation homotopy equivalences of paths. That is, if $[f]=[g]$ in the group then the homotopy $f\overset{\varphi}{\simeq}g$ necessarily has $\varphi(0,t)=x_0=\varphi(1,t)$ for all $t\in I$.
It is clear to me what (I think) the intended solution is:
Let $f$ be a loop about $x_0$ in $X$. Since $f(0)=f(1)$, $f$ can be identified as a continuous function $S^1\to X$. By assumption there is a homotopy $f\simeq C_{x}$ for some $x\in X$. Therefore the loop $f$ is nullhomotopic so $[f]$ is the identity in $\pi_1(X,x_0)$. As $f$ was arbitrary, $\pi_1(X,x_0)$ is trivial.
The problem is that this homotopy might not be a homotopy of paths, so $[f]=[C_{x_0}]$ might not - necessarily - hold. There is also, I'm pretty sure, nothing further we can conclude along the lines of: If such homotopies always exist, homotopies of paths must also always exist. So I think, as written, the exercise is wrong.
Is this an oversight, or have I misinterpreted what Hatcher was referring to? For context, this is the full exercise:
Let $X$ be a space. Show that all three are equivalent:
- All maps $f:S^1\to X$ are nullhomotopic
- All maps $f:S^1\to X$ can be extended to maps $f:D^2\to X$.
- $\pi_1(X,x_0)=0$ for all $x_0\in X$.
Now, I am sure that Hatcher was right to remark that we need to not assume homotopies preserve endpoints in $(1)\iff(2)$, since in fact (my solution for) $(2)\implies(1)$ doesn't guarantee a homotopy of paths. So perhaps he meant the remark for $(1)$ and $(2)$ but did not intend for it to apply to $(3)$, but that wouldn't really make sense. What's going on?
