Fake proof: All spaces are homotopically equivalent

Question

The definition of homotopy equivalence in Munkre's book:

An exercise from the same book:

My problem with this:

If all functions are homotopic, doesn't that mean that all spaces are homotopically equivalent?! In fact we get $g\circ f\simeq i_X$ and $f\circ g\simeq i_Y$ whatever functions $f,g$ we choose, and looking at the definition this means that $X,Y$ are homotopically equivalent, whatever they are. Where am I mistaken?

Answer 1

The mistake is that the exercise shows that any two maps to $\mathbb{R^n}$, specifically, are homotopic. It's a property of $\mathbb{R^n}$. What it means for homotopy equivalence is that, when we're trying to show $X$ is homotopy equivalent to $\mathbb{R^n}$, it suffices to build a the homotopy between our maps on the "$X$" side, since we automatically get one on the $\mathbb{R^n}$ side. A corollary is that $\mathbb{R^m}$ is homotopy equivalent to $\mathbb{R^n}$ for all $m,n$, which is true.

Answer 2

The result shown in the exercise allows us to show that $\Bbb R^n$ and $\Bbb R^m$ are homotopy equivalent, but cannot be applied in the general case: If we let $X=\Bbb R^n$ and $Y=\Bbb R^m$, it is easy to come up with some continuous (e.g., constant) maps $f\colon X\to Y$ and $g\colon Y\to X$. Then the continuous maps $g\circ f\colon X\to X=\Bbb R^n$ and $i_X$ are continuous maps form some space $X$ to $\Bbb R^n$, and by the exercise are homotopic. The same holds for $f\circ g$ and $i_Y$, as the exercise applies also to $Y=\Bbb R^m$. However, this trick wonÄt work if either of $X,Y$ is not some $\Bbb R^{\text{something}}$ ...