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 (\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

Example of simply connected space

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.

Lemma for simply connected space

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

Proof

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}

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.

Check the homomorphism property


\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}

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.

Proof

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}

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])

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