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How does one find a closed formula for the following?

If a space is simply-connected meaning that loops are null-homotopic, then arbitrary paths in the space are homotopic to one another using only the definition of homotopy?

Any suggestions are appreciated!

Edit: I made an attempt at a solution, which I will include tomorrow; the issue occurs when $t$, as described below, is equal to $1/2$. If anyone sees a way around this that some how keeps my overall idea, please let me know; I wasn't comfortable just using suggestions on here-I wanted to create something of my own, but I definitely used some of the suggestions for help!

Here's the situation:

Let $\Omega$ be a space that is simply-connected in the sense of loops. Let $\alpha, \beta$ be distinct points in $\Omega$. By assumption, $\Omega$ is also path-connected, and so there are paths $\gamma_1,\gamma_1 \colon I \to \Omega$, where $I = [0,1]$, that connect $\alpha$ to $\beta$ and $\beta$ to $\alpha$, respectively. Define their product $\gamma \colon I \to \Omega$, a loop with base point $\alpha$ by $$ \gamma(s) = \begin{cases} \gamma_1(2s) &\text{if } s \in [0,1/2]\\ \gamma_2(1-2s) &\text{ if } s \in [1/2,1]. \end{cases} $$ The space $\Omega$ is simply connected and so there is a point $z \in \Omega$ and a family of loop $\{\sigma_s(t)\}$ that continuous deforms $\gamma$ onto the point $z$. We want to find a family that takes the path $\gamma_1$ to $\gamma_2$ without moving the points $\alpha$ or $\beta$. Here is my suggested family $\{\tau_t(s)\}$, which I define in two pieces. When $0 \leq t < 1/2$, define $$ \tau_t(s) = \begin{cases} \sigma_s(0) &\text{if } s \in [0,t]\\ \sigma_t\left(\frac{s-t}{2(1-2t)}\right) &\text{ if } s \in [t,1-t]\\ \sigma_{1-s}(1/2) &\text{ if } s \in [1-t,1]\\ \end{cases} $$ and for $t \in [1/2,1]$, we have $$ \tau_t(s) = \begin{cases} \sigma_s(1/2) &\text{if } s \in [0,1-t]\\ \sigma_t\left(\frac{s-t}{2(1-2t)}\right) &\text{ if } s \in [1-t,t]\\ \sigma_{1-s}(0) &\text{ if } s \in [t,1]\\ \end{cases} $$

This appears to be a homotopy of paths that takes $\gamma_1$ onto $\gamma_2$, however, there is an issue when $t=1/2$. Does any one have any suggestions for adjusting this? :)

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    $\begingroup$ Perhaps prove that "being homotopic to" is an equivalence relation? (So if two loops are both homotopic to the trivial loop, then they are homotopic to each other...) $\endgroup$ – Kenny Wong Apr 11 '17 at 21:52
  • $\begingroup$ I don't know that that would help; I basically have to show that simple connectedness for loops implies simple connectedness for paths. I mean, my general idea was to take a path, deform it to a point using a homotopy, then follow the image of zero under the homotopy to time 1/2 to create a path, follow along the curve of the homotopy, connecting it to some other point on the original loop, then finally following the original homotopy, but I can't find a way to write this explicitly. $\endgroup$ – mathgenesis22813 Apr 11 '17 at 22:27
  • $\begingroup$ Suppose you have two paths sharing endpoints, which are not homotopic. What can you then say about the loop obtained by following one path followed by the other? $\endgroup$ – Joppy Apr 12 '17 at 1:49
  • $\begingroup$ Both the curves are homotopic to points. Simply-connected spaces are also required to be path-connected, so there is a curve connecting the two points. Take the homotopy of the first curve to a point, then follow by moving the point along the connecting curve. then end with the reverwe of the 2nd homotopy, expanding the 2nd point out to the 2nd curve. $\endgroup$ – Paul Sinclair Apr 12 '17 at 3:24
  • $\begingroup$ @PaulSinclair I actually do have a picture similar to that one, however, I haven't been able to find an actual explicit formula that cane make this picture work out. $\endgroup$ – mathgenesis22813 Apr 20 '17 at 18:03
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Let $X$ be the space, and $\gamma_1, \gamma_2 : I \to X$ be the two curves, where $I = [0,1]$. Because $X$ is simply-connected, we know that there are homotopies $h_1, h_2 : I^2 \to X$ between $\gamma_1, \gamma_2$ and points $p_1, p_2 \in X$, respectively. That is, for all $t\in I$ and $i = 1,2$,

  • $h_i(0,t) = \gamma_i(t)$ and $h_i(1,t) = p_i$

Also, since $X$ is simply-connected, it is path-connected, so there exists a curve $\beta : I \to X$ such that $\beta(0) = p_1$ and $\beta(1) = p_2$.

Define $h : I^2 \to X$ by: $$h(s, t) = \begin{cases}h_1(3s, t)& 0 \le s \le \frac 13\\ \beta(3s-1, t)& \frac 13 \le s \le \frac 23\\ h_2(3-3s, t)& \frac 23 \le s \le 1\end{cases}$$

Then $h$ is your desired homotopy.

As an aside, note that the first 2/3 of this homotopy proves that if $X$ is simply-connected, then any curve in $X$ is homoptopic to any point, not just some point.

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  • $\begingroup$ That makes sense! Thank you! $\endgroup$ – mathgenesis22813 Apr 22 '17 at 21:16
  • $\begingroup$ @John Don - thank you for the correction. $\endgroup$ – Paul Sinclair Nov 22 '17 at 23:42

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