Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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. – MathematicalPhysicist Nov 10 '11 at 11:54
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 $x_0$. – Joe Johnson 126 Nov 10 '11 at 13:48
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? – MathematicalPhysicist Nov 10 '11 at 17:33
Yes. That is correct. – Joe Johnson 126 Nov 10 '11 at 18:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.