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Math 344-1,2: Kursobjekt

Section 2.2 Fundamental group

Definition 2.2.1. Group compendium.

A group is a set \(G\) together with a binary operation \(\cdot G\times G\rightarrow G\) such that (i) the operation is associative, (ii) there is an identity element \(e\in G\) satisfying \(e\cdot g=g\cdot e=g\) for all \(g\in G\text{,}\) and (iii) for all \(g\in G\) there is an inverse element \(g^{-1}\in G\) satisfying \(g\cdot g^{-1}=g^{-1}\cdot g=e\text{.}\)
A subgroup of a group \(G\) is a subset \(H\subseteq G\) that contains the identity element \(e\in G\text{,}\) and which is closed under the group operation: i.e., if \(h, h'\in H\text{,}\) then \(h\cdot h'\in H\text{.}\) We write \(H\leq G\) in this case.
A normal subgroup of \(G\) is a subgroup \(N\) that satisfies \(gNg^{-1}=N\) for all \(g\in G\text{:}\) i.e., \(N\) is closed under conjugation. We write \(N\normalin G\) in this case.
Given a subgroup \(H\leq G\text{,}\) we denote by \(G/H\) the set of all left cosets of \(H\text{:}\) i.e., \(G/H=\{gH\colon g\in G\}\text{.}\) Given a normal subgroup \(N\normalin G\) the set of cosets \(G/N\) has a group structure given by the operation
\begin{equation*} gN\cdot g'N=gg'N\text{.} \end{equation*}
We call \(G/N\) the quotient of \(G\) by \(N\text{.}\)
A homomorphism between groups \(G, G'\) is a function \(\phi\colon G\rightarrow G'\) satisfying \(\phi(g_1\cdot_G g_2)=\phi(g_1)\cdot_{G'} \phi(g_2)\) for all \(g_1, g_2\in G\text{.}\) (Here we use subscript notation to differentiate the two different group operations at play.)
An isomorphism is an invertible homomorphism \(\phi\colon G\rightarrow G'\text{.}\) We say \(G\) and \(G'\) are isomorphic, denoted \(G\cong G'\text{,}\) if there is an isomorphism between them. (As it turns out, the inverse of an invertible group homomorphism is automatically a group homomorphism.)

Definition 2.2.2. Fundamental group.

Given a topological space \(X\) and element \(x_0\in X\text{,}\) the fundamental group of \(X\) based at \(x_0\text{,}\) denoted \(\pi_1(X,x_0)\text{,}\) is defined as
\begin{equation*} \pi_1(X,x_0)=P(X; x_0, x_0)/\simeq_p\text{.} \end{equation*}
In other words, \(\pi_1(X,x_0)\) is the set of all path homotopy equivalence classes \([f]\text{,}\) where \(f\in P(X; x_0, x_0)\) is a closed path (or loop) beginning and ending at \(x_0\text{.}\)
The path product operation
\begin{align*} \pi_1(X,x_0)\times \pi(X,x_0) \amp \rightarrow \pi_1(X,x_0)\\ ([f],[g]) \amp \mapsto [f]*[g]=[f*g] \end{align*}
endows \(\pi_1(X,x_0)\) with the structure of a group.

Identity element.

The identity element of \(\pi_1(X,x_0)\) is given by \([e_{x_0}]\text{,}\) where \(e_{x_0}\colon I\rightarrow X\) is the constant function \(e_{x_0}(s)=x_0\) for all \(s\in I\text{.}\)

Inverse elements.

Given an element \(g=[f]\in \pi_1(X,x_0)\) its group inverse is \(g^{-1}=[\overline{f}]\in \pi_1(X,x_0)\text{,}\) where \(\overline{f}\) is the reverse of \(f\text{.}\)

Proof.

That the path product operation endows \(\pi_1(X,x_0)\) with the structure of a group follows directly from Theorem 2.1.12.

Definition 2.2.3. Simply connected.

A topological space \(X\) is simply connected if it is path connected and \(\pi_1(X,x_0)=\{e\}\) is trivial for all \(x_0\in X\text{.}\)

Proof.

Note that the formula for \(\hat{\alpha}\) is defined in terms of the path product operation. We showed in Definition 2.1.11 that this operation is well-defined on equivalence classes, making \(\hat{\alpha}\) well defined.
We now show \(\hat{\alpha}\) is a homomorphism of groups. Given \([f], [g]\in \pi_1(X,x_0)\text{,}\) we have
\begin{align*} \hat{\alpha}([f]*[g]) \amp = [\overline{\alpha}]*[f]*[g]*[\alpha]\\ \amp =[\overline{\alpha}]*[f]*[e_{x_0}]*[g]*[\alpha]\\ \amp =[\overline{\alpha}]*[f]*[\alpha]*[\overline{\alpha}]*[g]*[\alpha]\\ \amp =\hat{\alpha}([f])*\hat{\alpha}([g])\text{.} \end{align*}
Lastly one easily shows that \(\widehat{\alpha}\) is invertible, with inverse \(\widehat{\alpha}^{-1}=\widehat{\overline{\alpha}}\text{.}\)

Definition 2.2.5. Pointed space.

Given a topological space \(X\) and element \(x_0\in X\text{,}\) the pair \((X,x_0)\) is called a pointed space with base point \(x_0\text{.}\) Given pointed spaces \((X,x_0)\) and \((Y,y_0)\text{,}\) a map of pointed spaces (or based map) is a continuous function \(f\colon X\rightarrow Y\) satisfying \(f(x_0)=y_0\text{.}\) We write \(f\colon (X,x_0)\rightarrow (Y,y_0)\) in this case.

Proof.

  1. It is clear that the rule \(f\mapsto h\circ f\) defines a map from \(P(X;x_0,x_0)\) to \(P(Y; y_0, y_0)\text{.}\) We must show that it is well defined on equivalence classes. Assume \(f,f'\in P(X; x_0, x_0)\) are path homotopic, and let \(F\) be a path homotopy from \(f\) to \(f'\text{.}\) It is easy to see that \(h\circ F\) is a path homotopy from \(h\circ f\) to \(h\circ f'\text{.}\) Thus \([f]=[f']\) implies \([h\circ f]=[h\circ f']\text{,}\) showing \(h_*\) is a well-defined map from \(\pi_1(X,x_0)\) to \(\pi_1(Y,y_0)\text{.}\)
    Now we show \(h_*\) is a homomorphism of groups. We will use the (easy to show fact) general fact that given \(f, g\in P(X; x_0, x_1)\) and continuous \(h\colon X\rightarrow Y\text{,}\) we have \(h\circ (f*g)=(h\circ f)*(h\cirg g)\text{.}\) Now, given \([f],[g]\in \pi_1(X, x_0)\text{,}\) we have
    \begin{align*} h_*([f]*[g]) \amp = [h\circ (f*g)]\\ \amp = [(h\circ f) * (h\circ g)\\ \amp =[h\circ f]*[h\circ g]\\ \amp =h_*([f])*h_*([g])\text{.} \end{align*}
  2. For any \(f\in \pi_1(X,x_0)\) we have
    \begin{align*} (k\circ h)_* ([f]) \amp= [(k\circ h)\circ f] \\ \amp = [k\circ (h\circ f)]\\ \amp =k_*([h\circ f])\\ \amp =k_*(h_*([f]))\\ \amp =k_*\circ h_*([f])\text{.} \end{align*}
    This proves \((k\circ h)_*=k_*\circ h_*\text{.}\)
  3. For any \([f]\in \pi_1(X,x_0)\) we have \(\id_*([f])=[\id\circ f]=[f]\text{.}\) This proves \(\id_*\) is the identity homomorphism on \(\pi_1(X, x_0)\text{.}\)

Proof.

If \(h\) is a homemorphism, then \(h^{-1}\) defines a map of pointed spaces from \((Y,y_0)\) to \((X,x_0)\text{.}\) Since \(h\circ h^{-1}=\id_Y\) and \(h^{-1}\circ h=\id_Y\) we have
\begin{align*} h_*\circ (h^{-1})_* \amp = (h\circ h^{-1})_* \amp (h^{-1})_*\circ h_* \amp = (h^{-1}\circ h)_* \\ \amp =(\id_Y)_* \amp \amp =(\id_X)_* \\ \amp =\id_{\pi_(Y,y_0)} \amp \amp =\id_{\pi_1(X,x_0)}\text{.} \end{align*}
This shows that \((h^{-1})_*\) is the inverse of \(h_*\text{,}\) and thus that \(h_*\) is an isomorphism.