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# 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.