# Math4201 Topology II (Lecture 12)

## Algebraic topology

### Fundamental group

Recall from last lecture, the $(pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(pi_1(X,x_0),*)$ is isomorphic to $(pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.

> [!TIP]
> 
> How does the $\hat{\alpha}$ (isomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?

#### Definition of simply connected

A space $X$ is simply connected if 

- $X$ is [path-connected](https://notenextra.trance-0.com/Math4201/Math4201_L23/#definition-of-path-connected-space) ($\forall x_0,x_1\in X$, there exists a continuous function $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$)
- $pi_1(X,x_0)$ is the trivial group for some $x_0\in X$

<details>
<summary>Example of simply connected space</summary>

Intervals are simply connected.

---

Any star-shaped is simply connected.

---

$S^1$ is not simply connected, but $n\geq 2$, then $S^n$ is simply connected.

</details>

#### Lemma for simply connected space

In a simply connected space $X$, and two paths having the same initial and final points are path homotopic.

<details>
<summary>Proof</summary>

Let $f,g$ be paths having the same initial and final points, then $f(0)=g(0)=x_0$ and $f(1)=g(1)=x_1$.

Therefore $[f]*[\bar{g}]\simeq_p [e_{x_0}]$ (by simply connected space assumption).

Then

$$
\begin{aligned}
[f]*[\bar{g}]&\simeq_p [e_{x_0}]\\
([f]*[\bar{g}])*[g]&\simeq_p [e_{x_0}]*[g]\\
[f]*([\bar{g}]*[g])&\simeq_p [e_{x_0}]*[g]\\
[f]*[e_{x_1}]&\simeq_p [e_{x_0}]*[g]\\
[f]&\simeq_p [g]
\end{aligned}
$$

</details>

#### Definition of group homomorphism induced by continuous map

Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.

$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.

<details>
<summary>Check the homomorphism property</summary>

$$
\begin{aligned}
h_*([f]*[g])&=h_*([f*g])\\
&=[h_*[f*g]]\\
&=[h_*[f]*h_*[g]]\\
&=[h_*[f]]*[h_*[g]]\\
&=h_*([f])*h_*([g])
\end{aligned}
$$

</details>

#### Theorem composite of group homomorphism

If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:pi_1(X,x_0)\to pi_1(Z,z_0)$ where $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$, $k_*:pi_1(Y,y_0)\to pi_1(Z,z_0)$,is a group homomorphism.

<details>
<summary>Proof</summary>

Let $f$ be a loop based at $x_0$.

$$
\begin{aligned}
k_*(h_*([f]))&=k_*([h\circ f])\\
&=[k\circ h\circ f]\\
&=[(k\circ h)\circ f]\\
&=(k\circ h)_*([f])\\
\end{aligned}
$$

</details>

#### Corollary of composite of group homomorphism

Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:pi_1(X,x_0)\to pi_1(X,x_0)$.

If $h$ is a homeomorphism with the inverse $k$, with

$$
k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
$$

This induced $h_*: pi_1(X,x_0)\to pi_1(Y,y_0)$ is an isomorphism.

#### Corollary for homotopy and group homomorphism

If $h,k:(X,x_0)\to (Y,y_0)$ are homotopic maps form $X$ to $Y$ such that the homotopy $H_t(x_0)=y_0,\forall t\in I$, then $h_*=k_*$.

$$
h_*([f])=[h\circ f]\simeq_p[k\circ h]=k_*([f])
$$
