# Math4202 Topology II (Lecture 28)

## Algebraic Topology

### Fundamental Groups of Some Surfaces

Recall from last week, we will see the fundamental group of $T^2=S^1\times S^1$, and $\mathbb{R}P^2$, Torus with genus $2$.

Some of them are abelian, and some are not.

#### Theorem for fundamental groups of product spaces

Let $X,Y$ be two manifolds. Then the fundamental group of $X\times Y$ is the direct product of their fundamental groups, 

i.e.

$$
\pi_1(X\times Y,(x_0,y_0))=\pi_1(X,x_0)\times \pi_1(Y,y_0)
$$

<details>
<summary>Proof</summary>

We need to find group homomorphism: $\phi:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(X,x_0)\times \pi_1(Y,y_0)$.

Let $P_x,P_y$ be the projection from $X\times Y$ to $X$ and $Y$ respectively.

$$
(P_x)_*:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(X,x_0)
$$

$$
(P_y)_*:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(Y,y_0)
$$

Given $\alpha\in \pi_1(X\times Y,(x_0,y_0))$, then $\phi(\alpha)=((P_x)_*\alpha,(P_y)_*\alpha)\in \pi_1(X,x_0)\times \pi_1(Y,y_0)$.

Since $(P_x)_*$ and $(P_y)_*$ are group homomorphism, so $\phi$ is a group homomorphism.

**Then we need to show that $\phi$ is bijective.** Then we have the isomorphism of fundamental groups.

To show $\phi$ is injective, then it is sufficient to show that $\ker(\phi)=\{e\}$.

Given $\alpha\in \ker(\phi)$, then $(P_x)_*\alpha=\{e_x\}$ and $(P_y)_*\alpha=\{e_y\}$, so we can find a path homotopy $P_X(\alpha)\simeq e_x$ and $P_Y(\alpha)\simeq e_y$.

So we can build $(H_x,H_y):X\times Y\times I\to X\times I$ by $(x,y,t)\mapsto (H_x(x,t),H_y(y,t))$ is a homotopy from $\alpha$ and $e_x\times e_y$.

So $[\alpha]=[(e_x\times e_y)]$. $\ker(\phi)=\{[(e_x\times e_y)]\}$.

Next, we show that $\phi$ is surjective.

Given $(\alpha,\beta)\in \pi_1(X,x_0)\times \pi_1(Y,y_0)$, then $(\alpha,\beta)$ is a loop in $X\times Y$ based at $(x_0,y_0)$. and $(P_x)_*([\alpha,\beta])=[\alpha]$ and $(P_y)_*([\alpha,\beta])=[\beta]$.
</details>

#### Corollary for fundamental groups of $T^2$

The fundamental group of $T^2=S^1\times S^1$ is $\mathbb{Z}\times \mathbb{Z}$.

#### Theorem for fundamental groups of $\mathbb{R}P^2$

$\mathbb{R}P^2$ is a compact 2-dimensional manifold with the universal covering space $S^2$ and a $2-1$ covering map $q:S^2\to \mathbb{R}P^2$.

#### Corollary for fundamental groups of $\mathbb{R}P^2$

$\pi_1(\mathbb{R}P^2)=\#q^{-1}(\{x_0\})=\{a,b\}=\mathbb{Z}/2\mathbb{Z}$

Using the path-lifting correspondence.

#### Lemma for The fundamental group of figure-8

The fundamental group of figure-8 is not abelian.