Show that if $X$ is path-connected, then every path $\alpha :I \to X$ are homotopic with each other (neglecting the condition rel $\{0,1\})$.

general-topologyhomotopy-theory

Show that if $X$ is path-connected, then every path $\alpha :I \to X$ are homotopic with each other (neglecting the condition rel $\{0,1\})$.

Since $X$ is path-connected we have that for every $a,b \in X$ there exists $\beta :I \to X$ such that $\beta(0)=a$ and $\beta(1)=b$.

Now for paths $\alpha, \gamma : I \to X$ what I need to find is a continuous map $h:I^2 \to X$ such that $$h(x,0)=\alpha(x) \text{ and } h(x,1)=\gamma(x).$$

How is the path-connectedness condition helpful here? I don't see how I can use it. I suppose I somehow need to take into account the fact that I can join every two points with paths when considering the homotopy, but I cannot figure out how.

Best Answer

To solve this problem you want to take advantage of a very important fact about homotopy, namely it is an equivalence relation.

Knowing that, suppose you can also prove the following two things:

  1. Every path is homotopic to a constant path (which is done already in another link on this site).
  2. Any two constant paths are homotopic to each other, assuming path connectivity (which you did in the comments).

So, let $f, g : [0,1] \to X$ be any two paths in a path connected space $X$.

You know from item 1 that there exists a constant path $P : [0,1] \to X$ such that

  1. $f$ is homotopic to $P$.

You also know from item 1 that $g$ is homotopic to some constant path that I'll denote $Q : [0,1] \to X$. Applying the symmetry law, it follows that

  1. $Q$ is homotopic to $g$.

You also know from item 2 that

  1. $P$ is homotopic to $Q$.

Since $f$ is homotopic to $P$, and $P$ is homotopic to $Q$, and $Q$ is homotopic to $g$, you may conclude from the transitive law that $f$ is homotopic to $g$.

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