How to show the fundamental group of torus is abelian in a homotopic way?

Question

I know the torus is homeomorphic to $S^1 \times S^1$ and the fundamental group is $ \mathbb{Z} \times \mathbb{Z} $, but in the real case, (let the generators of the torus's fundamental group be $a$ and $b$). Like in the case of $S^1 \vee S^1$ product $abbbbaaa$ can't be expressed by $ \mathbb{Z} \times \mathbb{Z} $, I feel in torus it's hard to find the homotopy between $abbbbbaaaa$ and $a^5b^4$, in other words why is the product $aba$ the same with $a^2b$?

Answer 1

Draw a picture.

If you know how to visualize the universal cover of $S^1 \times S^1$, then it will become very clear how to visualize the homotopy. The universal cover is $\mathbb{R}\times \mathbb{R}$. Each lift of the path $a$ is a horizontal path from $(m,n)$ to $(m+1,n)$ for some integers $m,n$. Similarly each lift of the path $b$ is a vertical path from $(m,n)$ to $(m,n+1)$ for some integers $m,n$.

So, each lift of $aba$ is a "horizontal-vertical-horizontal" path, and each lift of $a^2 b$ is a "horizontal horizontal vertical" path. If you start both of these lifts at the integer lattice point $(0,0)$ then they will both end at the same point $(2,1)$.

Now you have two paths in $\mathbb{R}\times\mathbb{R}$ with the same initial endpoint and the same terminal endpoint, and it should be pretty clear how to construct a homotopy between them in $\mathbb{R}\times\mathbb{R}$. In fact, you ought to be able to use the intuition gained from the picture to write down an actual formula for the homotopy.

And once you have done that, just project back down to the torus using the universal covering map $\mathbb{R}\times\mathbb{R} \to S^1 \times S^1$.

Answer 2

We just need to show a path homotopy between $ab$ and $ba$.

The mental block for me was thinking that since the path $ab$ stops at the base point halfway through, then every path in the homotopy from $ab$ to $ba$ must also make a pitstop at the base point.

But this isn't true, all we need is for the paths in the homotopy to start and end at the base point, which really frees up the curve visually on the torus.

Also, a very easy way to see the homotopy is on the square with sides identified appropriately, just like is mentioned here. If you pick your base point representative to be the bottom left vertex, $ab$ looks like traveling right first then up, while $ba$ travels up then right, which ends at the top right corner which is identified with the base point. The path homotopy is just sliding the path across the square.