### Racism and path homotopy

I recently had to face some racism on math.stackexchange.com for asking some seemingly obvious question on path homotopy. In retrospect, I could have thought through the concept myself. However, I don’t think I would have done so as quickly had I not been angry about the racism.

This is what I was trying to prove: let $\alpha$ be a path from $x_1$ to $x_2$. Then $\alpha*\alpha^{-1}*p*\alpha*\alpha^{-1}$ is homotopic to $p$. Here $p$ is a loop on $x_1$.

Obviously one might expect $\alpha$ and $\alpha^{-1}$ to cancel out on both sides, giving only $p$. However, I was interested in deriving it from first principles, and not just learn a seemingly obvious tool for future calculations. Hence the pedant fought for two days almost to finally understand this (turned out to be pretty elementary in the end).

You have a path homotopy between paths $f,g\in X$ when mapping $[0,1]\times\{0\}$ to $X$ gives you $f$, mapping $[0,1]\times \{1\}$ gives you $g$, and mapping $[0,1]\times \{t\}$ to $X$ gives you the paths “in between”. Let us call the continuous function from $k:I^2\to X$.

Now let us do something weird: let us construct two paths $m,n$ with the same terminal points in $I^2$. As $I^2$ is convex, the two paths are path homotopic. Hence, there exists a continuous mapping $a:I^2\to I^2$ such that $a([0,1]\times\{0\})=m$ and $a([0,1]\times\{1\})=n$.

Looking at the whole picture: this is what we have: a mapping $a:I^2\to I^2$ which is continuous, and another mapping $k:a(I^2)\to X$, which is continuous on $a(I^2)$ (as $k$ is continuous on $I^2$ and $a(I^2)$ is but a subspace of $I^2$). Hence, $(k\circ a):I^2\to X$ is continuous, which shows $k(m)$ is homotopic to $k(n)$.

Now we come to the most important part. Let $m$ be the path in $I^2$ such that $k(m)=p$. Also, let $n$ be the path in $I^2$ such that $k(n)=\alpha*\alpha^{-1}*p*\alpha*\alpha^{-1}$. They have the same terminal points. Why their terminal points can be made to be the same has terrific ideas behind it, and would require another post (it is not explained in Munkres, so watch out for that post). Anyway, as $(k\circ a):I^2\to X$ is continuous, it is easily seen that $p$ and $\alpha*\alpha^{-1}*p*\alpha*\alpha^{-1}$ are homotopic.