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I would like to prove the following basic fact related to the fundamental group:

A loop $\gamma : [0,1] \rightarrow X$ is null homotopic if and only if it can be extended to a continuous function of the disk $D^2$.

Can you tell me if this is correct:

I'll use the following fact (without proof):

$(*)$ If $X$ is a cell complex and $A$ is a subcomplex and $(X,A)$ has the homotopy extension property and $A$ is contractible and $q:X \rightarrow X/A$ is the projection into the quotient space then $q$ is a homotopy equivalence.

$\implies$ Let $\gamma $ be homotopic to the constant function $c$, i.e. there exists a homotopy $h_t: [0,1] \times [0,1] \rightarrow X$ with $h_0 = \gamma$ and $h_1 = x_0$. To make $h_t$ into a function from $D^2$ rather than $I \times I$ one can quotient $I \times I$ by the relation that identifies points in the set $A := \{0 \} \times I \cup I \times \{1\} \cup \{1\} \times I$ which is part of the border of the square $I \times I$. $I \times I / A$ is homeomorphic to $D^2$ and by $(*)$, homotopy equivalent to $I \times I$. Furthermore, its border after quotienting is homeomorphic to $S^1$. Now instead of viewing $\gamma $ as a function from $[0,1]$ with $\gamma(0) = \gamma(1) = x_0$ one can equivalently view it as a function from $S^1 = [0,1] / \{0,1\}$. If $q:I \times I \rightarrow I \times I / A$ is the quotient map then the function $h_t \circ q^{-1} : I \times I / A \rightarrow I \times I \rightarrow X$ is a function from $D^2$ to $X$ with the desired property that restricted to $S^1$ it coincides with $\gamma$.

$\Longleftarrow$ Let $\Gamma : D^2 \rightarrow X$ be a function that coincides with $\gamma$ on $S^1$. $D^2$ is contractible, i.e. there exists a homotopy $h_t : I \times D^2 \rightarrow X$ such that $h_0 = id_{D^2}$ and $h_1 = id_*$. Then $\Gamma \circ h_t$ is a homotopy $\gamma \simeq c$.

I would be very grateful if you not only could check this for mistakes but also tell me if there is a shorter or better way of doing something. Thanks!

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1 Answer 1

up vote 3 down vote accepted

For the forward direction, there is a much more direct proof - mainly, a homotopy between a loop and a constant is a map from the disc already. It's in the spirit of your proof, but doesn't require the homotopy extension theorem.

So, let $f:S^1\rightarrow X$ and assume $f$ is null homotopoic. Let $F(x,t)$ be a homotopy between $f$ and a constant map with $F(x,1) = f$ and $F(x,0) = x_0$.

Here's the trick - think of $(x,t)$ coordinates as polar coordinates on the disc where $x$ corresponds to $\theta$ and $t$ corresponds to $r$. Alternatively, the space $S^1\times [0,1]/$~ where we identify $S^1\times\{0\}$ to a point is clearly homeomorphic to the disc, and it's easy to see that since $F(x,0) = x_0$ independent of the $x$ value, that $F$ descends to map on $S^1\times [0,1]/$~.

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...Perhaps, it's worth mentioning, that a map $A\to X$ is null-homotopic iff it extends to $CA$, the cone of $A$ (proof is exactly the same). –  Grigory M Jun 24 '11 at 14:07

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