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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

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

1. Factoring:
   • Setup:
     • [ilmath]*:\Omega(X,b)\times\Omega(X,b)\rightarrow\Omega(X,b)[/ilmath] - the operation of loop concatenation - [ilmath]*:(\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)[/ilmath]

     through

     • [ilmath](p,p):\Omega(X,b)\times\Omega(X,b)\rightarrow\pi_1(X,b)\times\pi_1(X,b)[/ilmath] by [ilmath](p,p):(\ell_1,\ell_2)\mapsto(p(\ell_1),p(\ell_2))[/ilmath]
       • where [ilmath]p:\Omega(X,b)\rightarrow\pi_1(X,b)[/ilmath] is the canonical projection of the equivalence relation. As such we may say:
       • [ilmath](p,p)[/ilmath] is given by by [ilmath](p,p):(\ell_1,\ell_2)\mapsto([\ell_1],[\ell_2])[/ilmath] instead
     • We must show:
       • [ilmath]\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][/ilmath]^{[Note 1]}
   • Proof:
     • Let [ilmath]\ell_1,\ell_2,\ell_1',\ell_2'\in\Omega(X,b)[/ilmath] be given
       • Suppose that [ilmath]\neg([\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'])[/ilmath] 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 [ilmath][\ell_1]=[\ell_1']\wedge[\ell_2]=[\ell_2'][/ilmath] holds. We must show that in this case we have [ilmath][\ell_1*\ell_2]=[\ell_1'*\ell_2'][/ilmath]
         • By homotopy invariance of loop concatenation we see exactly the desired result
     • Since [ilmath]\ell_1,\ell_2,\ell_1'[/ilmath] and [ilmath]\ell_2'[/ilmath] were arbitrary this holds for all.
   • Conclusion
     • We obtain [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]
   • Thus the group operation is:
     • [ilmath][\ell_1]\overline{*}[\ell_2]:=[\ell_1*\ell_2][/ilmath]
2. Associativity of the operation [ilmath]\overline{*} [/ilmath]
3. Existence of an identity element in [ilmath](\pi_1(X,b),\overline{*})[/ilmath]
4. For each element of [ilmath]\pi_1(X,b)[/ilmath] the existence of an inverse element in [ilmath](\pi_1(X,b),\overline{*})[/ilmath]

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• Identity element
• Inverses
• Association

See Homotopy class for these properties

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TODO: Mond p30

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