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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.

MathematicalPhysicist
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  • 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? – ensbana Dec 28 '20 at 22:06

1 Answers1

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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.

J126
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    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 $x_0$. – J126 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. – J126 Nov 10 '11 at 18:21
  • Hey, I realize this is an old post, but hopefully someone’s still following it to answer some question. For the third implication, the relationship between $S^1 \times I$, $S^1 \times I / (S^1 \times {1})$ and $X$ looks like one where the universal property of the quotient could be invoked. So is it okay to say that, by the universal property of the quotient, there exists an unique continuous $g: S^1 \times I / (S^1 \times {1}) \to X$ such that $H = g \circ \Pi$ (where $H: S^1 \times I \to X$ is the homotopy, and $\Pi$ the quotient map), without saying anything else about $g$? – ensbana Dec 28 '20 at 21:55
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    @ensbana Yes, you only need to prove the existence of the extension. – J126 Dec 29 '20 at 13:35