Understanding the proof of fundamental group of H-space is abelian

Question

I'm trying to understand this part of wikipedia page on H-space:

The fundamental group of an H-space is abelian. To see this, let $X$ be an H-space with identity $e$ and let $f$ and $g$ be loops at $e$. Define a map $F: [0,1]×[0,1] → X$ by $F(a,b) = f(a)g(b).$ Then $F(a,0)$ $= F(a,1) = f(a)e$ is homotopic to $f$, and $F(0,b) = F(1,b) = eg(b)$ is homotopic to $g$. It is clear how to define a homotopy from $[f][g]$ to $[g][f].$

"$F(a,0)=F(a,1)=f(a)$ is homotopic to $f$" -- So $F(a,0)=F(a,1)$ is homotopic to the loop $f$? But they're just equal to $f$? What exactly is homotopic to $f$?

Same question with "$F(0,b)=F(1,b)=g(b)$ is homotopic to $g$".

And what's the homotopy from $[f][g]$ to $[g][f]$? Can any one flesh out the details? (If anyone could write the explicit map, that would really help. Thanks.)

Answer 1

That argument seems to rely on the weaker definition given on top of the Wikipedia page that amounts to:

An $H$-space is a topological space $X$ along with a map $\mu:X\times X\rightarrow X$ and an element $e$ such that the maps $x\mapsto \mu(x,e)$ and $x\mapsto \mu(e,x)$ are both homotopic to the identity map $X\rightarrow X$ through maps preserving the base-point $e$.

Basically, this definition weakens the property of $e$ being a two-sided identity to it being merely a identity when considered up to homotopy. Then, when the page considers the map $F(a,b)=f(a)g(b)$, it can only say that $F(a,0)=F(a,1)=f(a)\cdot e$ is homotopic to $f(a)$, even though it would be equal, as you suggest, if $e$ were a true identity.

The homotopy Wikipedia is referring to is as follows: Let $\gamma_1$ be the path along the square $[0,1]\times [0,1]$ that traces from $(0,0)$ to $(1,0)$ to $(1,1)$ along line segments and $\gamma_2$ be the path that traces from $(0,0)$ to $(0,1)$ to $(1,1)$. You can define a homotopy $H:[0,1]\times [0,1]\rightarrow [0,1]\times [0,1]$ between these two curves by linearly interpolating between them.

In the case where $e$ is a true identity, then $F\circ \gamma_1 = fg$ and $F\circ\gamma_2=gf$, so $F\circ H$ is a homotopy of $fg$ to $gf$. In the case where $e$ is merely an identity up to homotopy, $F\circ \gamma_1$ is only homotopic to $fg$ and $F\circ \gamma_2$ is only homotopic to $gf$, so you need to compose the homotopy $F\circ H$ with some other homotopies, but the argument is similar.