# Fundamental groups of products

Hey all! Today’s post is written by an awesome genius friend of mine, Nina Anikeeva. Without further ado…

We discuss what happens in terms of fundamental groups when we combine two spaces together in various ways. Unless otherwise stated, we will assume that our spaces are path-connected.

The Fundamental Group of a product

Let’s start with an example. Consider the two-dimensional torus $T^{2} = S^{1} \times S^{1}.$ What might its fundamental group be? We know that $\pi_1 (S^1) = \mathbb{Z}$ is generated by a loop going around the circle, so $\pi_1 (T^2)$ is probably generated by a loop $a$ going around the equator and a loop $b$ going around a meridian. Are there any relations between these loops? It’s not hard to see that $ab = ba$. One can also notice this by considering the torus as a quotient of a square. So it seems like

$\pi_1 (T^2) \cong \mathbb{Z}^2 \equiv \langle a, b \mid ab = ba\rangle.$

One might guess that the fundamental group of the $n$-torus $T^{2} = S^{1} \times \cdots \times S^{1}$ is

$\pi_1 (T^n) \cong \mathbb{Z}^n \equiv \langle a_1, \dots, a_n \mid [a_i, a_j]\rangle,$

generated by loops going around each factor. We show that this pattern holds in general:

$\pi_1 (X \times Y, (x_0, y_0)) \cong \pi_1 (X, x_0) \times \pi_1 (Y, y_0).$

The intuition is that the two factors don’t really interact with each other. This is the essence of the proof; to make it rigorous, recall that the product topology has the property that if one writes a continuous function $f: Z \rightarrow X \times Y$ as $f(z) = (f_1(z), f_2(z))$, then $f$ is continuous iff both $f_1$ and $f_2$ are. In particular, loop at $(x_0,y_0)$ is the same as a pair of loops at $x_0$ and $y_0$.

A homotopy of loops at $(x_0,y_0)$ is also therefore the same as a homotopy of loops (in $X$) at $x_0$ and (in $Y$) $y_0$. Thus, we consider the map $\pi_1 (X \times Y, (x_0, y_0)) \cong \pi_1 (X, x_0) \times \pi_1 (Y, y_0)$ given by $[f]\mapsto ([f_1],[f_2])$. The above discussion tells us that this is a bijection. Obviously, this is a group homomorphism, and, consequently, and isomorphism.

Examples

What else can we do with circles? The other obvious thing to try is to glue them together at a point. This is the wedge sum $S^1 \vee S^1.$ As before, the obvious guess is that $\pi_1 (S^1 \vee S^1)$ is generated by a loop $a$ going around the first circle and a loop $b$ going around the second. But this time, there shouldn’t be any relations between $a$ and $b$. If $F_n$ denotes the free group on $n$ letters, then our guess is

$\pi_1 (S^1 \vee S^1) \equiv F_2 \cong \langle a, b\mid \rangle.$

One might guess that

$\pi_1 (S^1 \vee \cdots \vee S^1) \equiv F_n \cong \langle a_1, \dots, a_n\mid \rangle.$

If $X$ and $Y$ are spaces in general with chosen basepoints $x_0$ and $y_0$, respectively, then one can define $X\vee Y$ to be the quotient space of the disjoint union $X\coprod Y$ by identifying the basepoints $x_0$ and $y_0$. More generally, it seems plausible that $\pi_1 (X \vee Y)$ consists of all strings of paths in $X$ and $Y$ with no relations between the paths in $X$ and the paths in $Y$: form something by concatenating $\pi_1 (X)$ and $\pi_1 (Y)$ without any relations. To be formal, the free product $G \ast H$ of two groups $G \equiv \langle S\mid R\rangle$ and $H \equiv \langle T\mid Q\rangle$ is presented by

$G \ast H \equiv \langle S \sqcup T\mid R \sqcup Q\rangle.$

So it seems likely that

$\pi_1 (X\vee Y) \cong \pi_1 (X) \ast \pi_1(Y).$

Here’s another example. Start with the Möbius band $M$. The boundary of $M$ is a circle, so we can glue a disc $D^2$ to it. Call the result $X$. What does $\pi_1 (X)$ look like? Well, $M$ is homotopy equivalent to $S^1$, so $\pi_1(M) \cong \mathbb{Z}$. We also know that $\pi_1(D^2) = 0$. So it seems like we can get $\pi_1 (X)$ by taking $\pi_1 (M)$ and destroying some stuff. In particular, the loops in $D^2$ should be trivial. Since the boundary component of $M$ is represented by twice the generator, we see that

$\pi_1 (X) \equiv \langle a\mid a^2\rangle \cong \mathbb{Z}/2\mathbb{Z}$

You should convince yourself that $X$ is homeomorphic to $\mathbb{RP}^2$. If we view $\mathbb{RP}^2$ as a quotient of a sphere, then the nontrivial element of $\pi_1(\mathbb{RP}^2)$ is represented by a path going from the North pole to the South pole.

What happens if we glue together two Möbius bands together along their boundaries? Call this space $Y$. As above, ot seems like $\pi_1 (Y)$ is generated by a loop $a$ corresponding to a generator of the first band and a loop $b$ corresponding to a generator of the other. But these loops are related. Since the boundaries of the two Möbius bands have been identified, $a^2 = b^2$. Hence,

$\pi_1 (Y) \equiv \langle a, b\mid a^2 = b^2\rangle \cong \langle a, c\mid aca^{-1} = c^{-1}\rangle.$

The second presentation is obtained from the first by setting $c = ab^{-1}$ and exhibits $\pi_1 (Y)$ as a semi-direct product. It turns out that $Y$ is homeomorphic to the Klein bottle $K$. This is easily seen by drawing some pictures.

That concludes part I! Until next time,

Have fun,
SKD