I want your advice on my solution of this problem.
problem
Show that for a space $X$, the following three are equivalent:
(a) Every map $S^1 \rightarrow X$ is homotopic to a constant map, with image a point.
(b) Every map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$.
(c) $\pi_1(X,x_0)=0$ $\forall x_0 \in X$.
my solution
$(c) \rightarrow (a)$: Take a map $f:S^1\rightarrow X$, and the constant loop at $x_0$ (which from (c) we know there's only this loop), and take the linear homotopy connecting $f$ and $x_0$.
$(b)\rightarrow (c)$: I am not sure if this is right, but I want somehow to induce some isomorphism from $\pi_1(D^2,y_0)$ to $\pi_1(X,x_0)$ in which case we'll get the required result.
So I argued that we take as a map the extension of a map $S^1\rightarrow X$ to a map $D^2\rightarrow X$, but I don't see how to argue that the last map is homeomorphism, any advice?
$(a) \rightarrow (b)$: I am given $f:S^1\rightarrow X$, and it's homotopic to some point $x_0$, so I thought to extend $f$ on the whole ball, by defining the map:
$g(x)= f(x)$ if $|x|=1$ and $g(x)=x_0$ if $|x|<1$.
Is any of what I typed right, what do I need to amend?
Thanks.
Best Answer
For the third implication you take your homotopy $S^1\times I\rightarrow X$ that is constant at $S^1\times\{1\}$ and show that this map factors through the quotient of $S^1\times I$ by $S^1\times\{1\}$, which is homeomorphic to $D^2$.
For the second implication, you don't have $D^2\rightarrow X$ is a homeomorphism. The two sphere, $S^2$, has trivial $\pi_1$ but is not homeomorphic to $D^2$. Let $\alpha\in\pi_1(X,x_0)$. Then $\alpha$ has a representative $f_\alpha:S^1\rightarrow X$. By (b), this map extends to a map $D^2\rightarrow X$. This is the same as a map $S^1\times I\rightarrow X$ such that the restriction $S^1\times\{1\}\rightarrow X$ is a constant map. You just have to make sure this homotopy can be based.