Proof that the fundamental group is actually a group

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• In this article $I:=[0,1]\subset\mathbb{R}$ - the closed unit interval

Statement

Let $(X,\mathcal{ J })$ be a topological space and let $b\in X$ be given. Then $\Omega(X,b)$ denotes the "set of all loops based at $b$ in $X$". We claim that from this we can make a group, $(\pi(X,b),*)$ called the fundamental group where:

• $$\pi(X,b):=\frac{\Omega(X,b)}{\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\}\big)}$$
  • Where $\big({\small(\cdot)}\simeq{\small(\cdot)}\ (\text{rel }\{0,1\}\big)$ is the relation of end-point-preserving-homotopy on $C([0,1],X)$ - the space of all paths in $X$
    • In this case considered on $\Omega(X,b)$ which is a subset of $C(I,X)$.

TODO: Finish writing statement, mention that we're trying to factor loop concatenation through

Note: for an outline of the proof see below: outline

Proof

We wish to show that the set $\pi_1(X,b):=\frac{\Omega(X,b)}{\big({\small(\cdot)}\simeq{\small (\cdot)}\ (\text{rel }\{0,1\})\big)}$ is actually a group with the operation $\overline{*}$ as described in the outline.

1. Factoring:
   • Setup:
     • $*:\Omega(X,b)\times\Omega(X,b)\rightarrow\Omega(X,b)$ - the operation of loop concatenation - $*:(\ell_1,\ell_2)\mapsto\left(\ell_1*\ell_2:I\rightarrow X\text{ by }\ell_1*\ell_2:t\mapsto\left\{\begin{array}{lr}\ell_1(2t) & \text{for }t\in[0,\frac{1}{2}]\\ \ell(2t-1) & \text{for }t\in[\frac{1}{2},1]\end{array}\right.\right)$

     through

     • $(p,p):\Omega(X,b)\times\Omega(X,b)\rightarrow\pi_1(X,b)\times\pi_1(X,b)$ by $(p,p):(\ell_1,\ell_2)\mapsto(p(\ell_1),p(\ell_2))$
       • where $p:\Omega(X,b)\rightarrow\pi_1(X,b)$ is the canonical projection of the equivalence relation. As such we may say:
       • $(p,p)$ is given by by $(p,p):(\ell_1,\ell_2)\mapsto([\ell_1],[\ell_2])$ instead
     • We must show:
       • $\forall\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)\left[\big([\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2']\big)\implies\big([\ell_1*\ell_2]=[\ell_1'*\ell_2']\big)\right]$
   • Proof:
     • Let $\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)$ be given
       • Suppose that $\neg([\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'])$ holds, then by the nature of logical implication we're done, as we do not care about the RHS's truth or falsity in this case
       • Suppose that $[\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2']$ holds. We must show that in this case we have $[\ell_1*\ell_2]=[\ell_1'*\ell_2']$
         • By homotopy invariance of loop concatenation we see exactly the desired result
     • Since $\ell_1,\ell_2,\ell_1'$ and $\ell_2'$ were arbitrary this holds for all.
   • Conclusion
     • We obtain $\overline{*}:\pi_1(X,b)\times\pi_1(X,b)\rightarrow\pi_1(X,b)$ given unambiguously by:
       • $\overline{*}:([\ell_1],[\ell_2])\mapsto[\ell_1*\ell_2]$
   • Thus the group operation is:
     • $[\ell_1]\overline{*}[\ell_2]:=[\ell_1*\ell_2]$
2. Associativity of the operation $\overline{*}$
3. Existence of an identity element in $(\pi_1(X,b),\overline{*})$
4. For each element of $\pi_1(X,b)$ the existence of an inverse element in $(\pi_1(X,b),\overline{*})$

Grade: A*

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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