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The following are equivalent for a topological space X according to a problem in Hatcher.

$1$)Every continuous map $S^1 \to X$ is homotopic to a constant map.

$2$)Every continuous map $S^1 \to X$ extends to a continuous map $D^2 \to X$.

$3$)$\pi_1(X,x_0)=0$ for every $x_0 \in X$.

I am examining the proof of $2$) implies $3$).
Proof- Let $h:S^1 \to X$, then there is an extension $k:D^2 \to X$. If we let $j$ be the inclusion $S^1 \to D^2$, then $h=k \circ j$ and $h_*=k_* \circ j_*$. Since $\pi_1(D^2,s_0)=0$ for each $s_0 \in S^1$, then h* sends each element of $\pi_1(S^1,s_0)$ to the identity in $\pi_1(X,h(s_0))$.

So I have two issues. First, this only shows $h_*$ is the trivial homomorphism not $\pi_1(X,h(s_0))=0$. Second, even if I fix that then I seem to have only proved it for $x_0$ in the image of $h$. So if someone could help me with this I would be grateful.

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Is this a proof you wrote or a proof that Hatcher wrote? – Henry T. Horton Aug 12 '13 at 22:09
I wrote it not Hatcher; its a question in Hatcher – Leo Spencer Aug 12 '13 at 22:11
What you've done so far is on the right track. Hint: let $\alpha \in \pi_1(X,x_0)$. Then $\alpha$ is a class of maps $(S^1,s_0) \to (X,x_0)$. Pick one, call it $h$, then use your observation above to show that $\alpha = 0$. – Thomas Belulovich Aug 12 '13 at 22:18
@ThomasBelulovich $\alpha$ is a class of loops based at $x_0$. But these loops go from $I$ to $X$ not $S^1$ to $X$? – Leo Spencer Aug 12 '13 at 22:38
A map $f : I \to X$ satisfying $f(0) = f(1)$ gives a map $I \to S^1 \to X$. The circle is a quotient of $I$ gotten by gluing the ends together. – Thomas Belulovich Aug 12 '13 at 23:15
up vote 3 down vote accepted

Hint: A nullhomotopy $F:\mathbb{S}^1\times [0,1]\to X$ is equivalent to a map $G:\mathbb{D}^2\to X$.

I hope this hint is useful! (I can give you a hint for the hint if you'd like!)

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What does equivalent mean here? Homotopic? also, I do not see the relevance here to 2 implies 3. I used that essentially for 1 implies 2 – Leo Spencer Aug 12 '13 at 22:36
Dear @Leo, I mean that if I have a nullhomotopy $F:\mathbb{S}^1\times [0,1]\to X$, then I have a map $G:\mathbb{D}^2\to X$ and vice-versa. The point is that if you're given a map $\mathbb{S}^1\to X$ that extends to a map $\mathbb{D}^2\to X$, then you're just being given a null-homotopy of the original $\mathbb{S}^1\to X$. (Can you see why this is true?) Therefore, every (pointed) map $\mathbb{S}^1\to X$ is null-homotopic and $\pi_1(X,x_0)=0$. – Amitesh Datta Aug 12 '13 at 22:39
it makes sense that the what you are saying about the maps being null homotopic. However, I don't see why that means that the fundamental group is trivial. It seems that using 1 to imply 3. – Leo Spencer Aug 12 '13 at 23:15
@Leo Let $f:\mathbb{S}^1\to X$ be a (pointed) map. We wish to show (3) (that $f$ is null-homotopic) using (2). By (2), we know that there exists a map $g:\mathbb{D}^2\to X$ such that $g(x)=f(x)$ for all $x\in \mathbb{S}^1$. Let's define $H:\mathbb{S}^1\times [0,1]\to X$ by the rule $H(x,t)=g(tx)$. $H$ is the null-homotopy of $f:\mathbb{S}^1\to X$ which we seek! – Amitesh Datta Aug 13 '13 at 4:47
@Leo Note that (1) and (3) are obviously equivalent (they say exactly the same thing once you unwind the definition of the fundamental group). Also, the proof I've provided of (2)$\implies$(3) (or (2)$\implies$(1); they're the same thing to me) is an "if and only if proof" so it works in the other direction as well. (To say a map $f:\mathbb{S}^1\to X$ is null-homotopic is exactly the same thing as saying that it admits an extension to a map $g:\mathbb{D}^2\to X$ so there's nothing really to prove here. I think you've already observed this before asking the question as you stated.) – Amitesh Datta Aug 13 '13 at 4:50

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