Proof that the fundamental group is actually a group/Outline

• See Proof that the fundamental group is actually a group if you end up on this page looking for a proof. This is a subpage.

Outline of proof

[ilmath]\xymatrix{ \Omega(X,b)\times\Omega(X,b) \ar@2{->}[d]_{(\pi,\pi)} \ar[rr]^-{*} \ar@/^3.5ex/[drr]^(.75){\pi\circ *} & & \Omega(X,b) \ar[d]^{\pi} \\ \pi_1(X,b)\times\pi_1(X,b) \ar@{.>}[rr]_-{\overline{*} } & & \pi_1(X,b) }[/ilmath]
Factoring [ilmath]*[/ilmath] (loop concatenation) setup

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]b\in X[/ilmath] be given. Then [ilmath]\Omega(X,b)[/ilmath] is the set of all loops based at [ilmath]b[/ilmath]. Let [ilmath]{\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})[/ilmath] denote the relation of end-point-preserving homotopy on [ilmath]C([0,1],X)[/ilmath] - the set of all paths in [ilmath]X[/ilmath] - but considered only on the subset of [ilmath]C([0,1],X)[/ilmath], [ilmath]\Omega(X,b)[/ilmath].

Then we define: [math]\pi_1(X,b):=\frac{\Omega(X,b)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})\big)}[/math], a standard quotient by an equivalence relation.

Consider the binary function: [ilmath]*:\Omega(X,b)\times\Omega(X,b)\rightarrow\Omega(X,b)[/ilmath] defined by loop concatenation, or explicitly:

• [ilmath]*:(\ell_1,\ell_2)\mapsto\left(\ell_1*\ell_2:[0,1]\rightarrow X\text{ given by }\ell_1*\ell_2:t\mapsto\left\{\begin{array}{lr}\ell_1(2t) & \text{for }t\in[0,\frac{1}{2}]\\ \ell_2(2t-1) & \text{for }t\in[\frac{1}{2},1]\end{array}\right.\right)[/ilmath]
  • notice that [ilmath]t=\frac{1}{2}[/ilmath] is in both parts, this is a nod to the pasting lemma

We now have the situation on the right. Note that:

• [ilmath](\pi,\pi):\Omega(X,b)\times\Omega(X,b)\rightarrow\pi_1(X,b)\times\pi_1(X,b)[/ilmath] is just [ilmath]\pi[/ilmath] applied to an ordered pair, [ilmath](\pi,\pi):(\ell_1,\ell_2)\mapsto([\ell_1],[\ell_2])[/ilmath]

In order to factor [ilmath](\pi\circ *)[/ilmath] through [ilmath](\pi,\pi)[/ilmath] we require (as per the factor (function) page) that:

• [ilmath]\forall(\ell_1,\ell_2),(\ell_1',\ell_2')\in\Omega(X,b)\times\Omega(X,b)\big[\big((\pi,\pi)(\ell_1,\ell_2)=(\pi,\pi)(\ell_1',\ell_2')\big)\implies\big(\pi(\ell_1*\ell_2)=\pi(\ell_1'*\ell_2')\big)\big][/ilmath], this can be written better using our standard notation:
  • [ilmath]\forall\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)\big[\big(([\ell_1],[\ell_2])=([\ell_1'],[\ell_2'])\big)\implies\big([\ell_1*\ell_2]=[\ell_1'*\ell_2']\big)\big][/ilmath]

Then we get (just by applying the function factorisation theorem):

• [ilmath]\overline{*}:\pi_1(X,b)\times\pi_1(X,b)\rightarrow\pi_1(X,b)[/ilmath] given (unambiguously) by [ilmath]\overline{*}:([\ell_1],[\ell_2])\mapsto[\ell_1*\ell_2][/ilmath] or written more nicely as:
  • [ilmath][\ell_1]\overline{*}[\ell_2]:=[\ell_1*\ell_2][/ilmath]

Lastly we show [ilmath](\pi_1(X,b),\overline{*})[/ilmath] forms a group