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I am working on Hatcher's problem 1.1.5.

Show that for a space $X$, the following three conditions are equivalent.

$\textit{a)}$ Every map $S^1\to X$ is homotopic to a constant map.

$\textit{b})$ Every map $S^1\to X$ extends to a map $D^2\to X$

$\textit {c})$ $\pi_1(X,x_0)=0$.

I have shown that $a$ implies $b$.

Now I want to prove that $b$ implies $c$. Therefore, I let $[f]\in\pi_1(X,x_0)$. Then, $f:I\to X$ with $f(0)=f(1)=x_0$. I want to show that $f$ is homotopic to the constant path. Since $S^1\cong I/_{0\sim 1}$, $f$ induces a map $f':S^1\to X$ with $f'(1)=x_0$. By $b$, this map can be extended to a map $\tilde{f'}:D^2\to X$. Since $D^2$ is convex, $D^2$ is contractible, there exists a homotopy $\tilde{H}:I\times D^2\to X$, $\tilde{H}_0=const_{x_0}$, $\tilde{H}_1=Id_{D^2}$. I don't know how I can go on from this, There is also a point where I have to transfer back to $f$ without $'$.

For $c$ implies $a$: Let $[f]\in\pi_1(X,x_0)$. Then $f:I\to X$ is homotopic to the constant path $const_{x_0}$. Also here, I don't know how to go on.


marked as duplicate by Lee Mosher, Xander Henderson, José Carlos Santos, positrón0802, Cesareo Feb 18 at 19:43

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  • $\begingroup$ That question unfortunately does not answer my question, I have looked at it before. $\endgroup$ – user408856 Feb 13 at 20:08

Hint: If $f':S^1 \to X$ extends to a map $\tilde f: D^2 \to X$, this means $f'$ can be written as a composition $$ (S^1,1) \xrightarrow i (D^2,1) \xrightarrow {\tilde f} (X,x_0) $$ where $i$ is the inclusion. Can you find a homotopy $H_t$ rel $1$ from $i$ to some constant map $c$? What if you compose it with $\tilde f$?

For (c) $\implies$ (a), given a map $g:S^1\to X$, can you find a homotopy from $g$ to a loop at $x_0$?

  • $\begingroup$ What do you mean with, "$H_t$ rel 1 from i to some constant map c? And with $(S^1,1)$ the circle whereby 1 is mapped to $x_0$? $\endgroup$ – user408856 Feb 14 at 7:11
  • $\begingroup$ @James: I mean a map $H: S^1\times I \to D^2$ such that $H(x,0)=i(x)$ and $H(x,1)=1$ and $H(1,t)=1$ for all $t\in I, x\in S^1$. So the homotopy starts with the embedding $i$ and ends with a constant map, and is fixed at the point $1$ during the time interval. $\endgroup$ – Stefan Hamcke Feb 14 at 13:34
  • $\begingroup$ Since $D^2$ is convex, the homotopy $H(x,t)=(1-t)\cdot i(x)+t\cdot(1,0)$ will do? $\endgroup$ – user408856 Feb 16 at 9:40