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.

  • 2
    $\begingroup$ 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
  • 5
    $\begingroup$ "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


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.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .