I want your advice on my solution of this 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?


  • $\begingroup$ I’m doing a similar question, the only difference is that (a) is stated as:”Every closed path in $X$ is homotopic (with fixed endpoints) to a constant path”. But isn’t that just the definition of the trivial fundamental group? $\endgroup$ – ensbana Dec 28 '20 at 22:06

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.

  • 1
    $\begingroup$ I don't understand what does "The $\alpha$ has a representative $f_{\alpha}:S^1\rightarrow X$", means here. I mean $\alpha \in \pi_1(X,x_0)$ means that $\alpha$ is a homotopy class of loops based at $x_0$, shouldn't $f_{\alpha}$'s domain be from the unit interval? Thanks. $\endgroup$ – MathematicalPhysicist Nov 10 '11 at 11:54
  • $\begingroup$ Since $\alpha$ is a homotopy class, there is at least one map that represents it $f:I\rightarrow X$. Since we demand that $f(0)=f(1)=x_0$, this is the same as a map $f:S^1\rightarrow X$ such that the basepoint of $S^1$ maps to $x_0$. $\endgroup$ – Joe Johnson 126 Nov 10 '11 at 13:48
  • $\begingroup$ You mean we can look at it as a paramaterization from the interval $[0,2\pi]\rightarrow S^1 \rightarrow X$, so just take the composition of them, right? $\endgroup$ – MathematicalPhysicist Nov 10 '11 at 17:33
  • $\begingroup$ Yes. That is correct. $\endgroup$ – Joe Johnson 126 Nov 10 '11 at 18:21
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    $\begingroup$ @ensbana Yes, you only need to prove the existence of the extension. $\endgroup$ – Joe Johnson 126 Dec 29 '20 at 13:35

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