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

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

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

Then we define:

$$\pi_1(X,b):=\frac{\Omega(X,b)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\})\big)}$$ , a standard quotient by an equivalence relation.

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

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

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

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

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

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

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

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

Lastly we show $(\pi_1(X,b),\overline{*})$ forms a group