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It is clear to me that if all paths (with the same endpoints) in a region are homotopic then that region is simply connected, however I am having difficultly proving the converse, that is, all paths with the same endpoints are homotopic in a simply connected region. Here is what I have so far


Given two paths $\alpha$ and $\beta$ which begin and end at the same point, we can define a loop at either of those points by \begin{align} \gamma(t) = \begin{cases} \alpha(2t) &\quad 0 \leq t \leq 1/2\\ \beta(1-2t) &\quad 1/2 \leq t \leq 1\\ \end{cases} \end{align}

By assumption, the space is simply connected, and so all loops can be deformed into all other loops. The loop $\gamma$ can then be deformed to

\begin{align} \gamma'(t) = \begin{cases} \alpha(2t) &\quad 0 \leq t \leq 1/2\\ \alpha(1-2t) &\quad 1/2 \leq t \leq 1\\ \end{cases} \end{align}


Intuitively, it seems like there should then be a homotopy between $\alpha$ and $\beta$, however I cannot think of a rigorous way to show that one exists.

Any help would be appreciated, thanks.

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    $\begingroup$ What is you definition of "simply connected"? $\endgroup$ Oct 10, 2015 at 1:57
  • $\begingroup$ @MarianoSuárez-Alvarez That any loop can be continuously shrunk to a point. Are there other ways to define it? $\endgroup$
    – Haz
    Oct 10, 2015 at 2:00
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    $\begingroup$ "All paths are homotopic" is a sloppy setence. You mean any two paths with the same endpoints are homotopic by a homotopy with all intermediate paths going from the same points? In any path connected space, all paths - maps $[0,1]\to X$ - are homotopic, because they are all contractible to a constant map, and then path-connected means all constant maps are homotopic. $\endgroup$ Oct 10, 2015 at 2:01
  • $\begingroup$ @ThomasAndrews Yes I do, I will edit the question. $\endgroup$
    – Haz
    Oct 10, 2015 at 2:03

2 Answers 2

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There is also a purely algebraic solution. If the paths $\alpha$ and $\beta$ both start at $x$ and end at $y$, then

$[\alpha] = [\alpha * e_y] = [\alpha * (\bar{\beta} * \beta)]=[(\alpha * \bar{\beta}) * \beta] = [e_{x} * \beta] = [\beta]$,

where $e_x$ and $e_y$ denote the respective constant paths.

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Hint: Let $S^1$ be the unit circle and $D$ be the disk $|z|\leq 1$. Then show that if a map $f:S^1\to X$ is homotopic to a constant map, then there is a map $D\to X$ that agrees with $S^1\subset D$.

Once you have this $D\to X$, you can easily see that any two paths with the same end-points in $D$ are homotopic, because $D$ is convex, so $\gamma_1(x)(1-t)+\gamma_2(x)t$ is a homotopy between $\gamma_1$ and $\gamma_2$. If $\gamma_1$ and $\gamma_2$ have the same endpoints, this map has the same endpoints for all $t$.

In particular, in your example, the map from $\alpha$ and $\beta$ put together become a loop, and then the two paths are represented by the two ways around the circle to the midpoint.

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