## The Groupoid Properties of Operation * on Path-homotopy Classes (Proof)

Theorem: The operation * has the following properties:

(1) (Associativity) [f]*([g]*[h])=([f]*[g])*[h], i.e. it doesn’t matter where we place the brackets.

(2) (Right and left identities) Given $x\in X$, let $e_x$ denote the constant path $e_x: I\to X$ mpping all of I to the point x. If f is a path in X from $x_0$ to $x_1$, then $[f]*[e_{x_1}]=[f]$ and $[e_{x_0}]*[f]=[f]$.

(3) (Inverse) Given the path f in X from $x_0$ to $x_1$, let $\bar{f}$ be the path defined by $\bar{f}=f(1-s)$. $\bar{f}$ is called the reverse of f. Then, $[f]*[\bar{f}]=[e_{x_0}]$ and $[\bar{f}]*[f]=[e_{x_1}]$.

We will prove the above statements, of which (1) Associativity is actually the trickiest.

Proof:

We shall prove two elementary lemmas first. (This part is not proved in the book by Munkres).

Lemma 1: If $k: X\to Y$ is a continuous map, and if F is a path homotopy in X between the paths f and f’, then $k\circ F$ is a path homotopy in Y between the paths $k\circ f$ and $k\circ f'$.

Proof of Lemma 1: Since F is a path homotopy in X between paths f and f’, we have by definition that F(s,0)=f(s), F(s,1)=f'(s), F(0,t)=x_0, F(1,t)=x_1.

Then, k F(s,0)=kf(s), kF(s,1)=kf'(s), kF(0,t)=k(x_0), kF(1,t)=k(x_1). Since kF is continuous (composition of two continuous functions), kF is inded a path homotopy in Y between he paths kf and kf’.

Lemma 2: If $k:X\to Y$ is a continuous map and if f and g are paths in X with f(1)=g(0), then

$k\circ (f*g)=(k\circ f)*(k \circ g)$

Proof of Lemma 2:

$k(f*g)(s)=kh(s)$, where h=f*g as defined previously.

$(kf)*(kg)(s)=kh(s)$.

We will first verify property (2) on Right and Left Identities. Let $e_{x_0}$ denote the constant path in I at 0, and we let $i: I\to I$ denote the identity map, which is a path in I from 0 to 1. Then $e_0 * i$ is also a path in I from 0 to 1.

Because I is convex, there is a path homotopy G in I between i and $e_0 *i$ (Straight-line homotopy) Then $f\circ G$ is a path homotopy in X between the paths $f\circ i=f$ and $f\circ (e_0 *i)$ (Lemma 1). Furthermore by Lemma 2, $f\circ (e_0 *i) = (f \circ e_0) * (f \circ i)$ which is equivalent to $e_{x_0} *f$.

A similar argument, using the fact that if $e_1$ denotes the constant path at 1, then $i*e_i$ is path homotopic in I to the path i, shows that $[f]*[e_{x_1}]=[f]$.

To prove (3) (Inverse), we note that the reverse of i is $\bar{i}(s)=1-s$. Then $i*\bar{i}$ is a path in I beginning and ending at 0. The constant path $e_0$ is also beginning and ending at 0. Again, because I is convex, there is a path homotopy H in I between $e_0$ and $i*\bar{i}$ (straight-line homotopy). Then, using lemma 1 and 2, $f\circ H$ is path homotopy between $f\circ e_0=e_{x_0}$ and $f\circ (i*\bar{i})=(f\circ i)*(f\circ\bar{i})=f*\bar{f}$. Very similarly, we can use the fact that $\bar{i}*i$ is path homotopic  in I to $e_1$ to show that $[\bar{f}]*[f]=[e_{x_1}]$.

We will continue the proof of associativity (which is longer) in the next blog post.