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

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 $a^2b$?

• In general, $\pi_1(X \times Y, (x_0, y_0)) \cong \pi_1(X, x_0) \times \pi_1(Y, y_0)$. – user98602 Aug 21 '14 at 5:55
• It all begins (and in a way ends) with finding a homotopy between $ab$ and $ba$. Once you have that, then everything else follows from associativity. For example $$abbabb=ab(ba)bb=ab(ab)bb=ababbb=a(ba)bbb=a(ab)bbb=a^2b^4.$$ And finding that single homotopy is easy when you construct the torus by glueing pairs of opposite sides of a square in the appropriate way- just gradually move the path across the square. In $S^1\vee S^1$ that square is empty, and you cannot move a path across it. – Jyrki Lahtonen Aug 21 '14 at 6:02
• Thank you very much Jyrki! – 6666 Aug 21 '14 at 6:45

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 $\mathbb{R}\times\mathbb{R} \to S^1 \times S^1$.