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I'm trying to prove that the product operation on paths induces a well defined operation on path homotopy classes defined by the equation $[f]*[g]=[f*g]$

Let $F$ be a path homotopy between $f$ and $f'$. Then $F(s,0)=f$, $F(s,1)=f'$,
$F(0,t)=x_0$ and $F(1,t)=x_1$

Let $G$ be a path homotopy between $g$ and $g'$. Then $G(s,0)=g$, $G(s,1)=g'$ $G(0,t)=x_1$ and $G(1,t)=x_2$.

I want to show that $[f]*[g]=[f']*[g']$.

Define $H:[0,1]\times [0,1]\to X$ by $H(s,t)=F(2s,t)$ if $s$ in $[0,1/2]$ and
$G(2s-1,t)$ if $s$ in $[1/2,1]$

Then $H(1/2,t)=F(1,t)=x_1=G(0,t)$. Thus $H$ is well defined.

$F$ is cts on $[0,1/2]\times [0,1]$ and $G$ is cts on $[1/2,1]\times [0,1]$ thus $H$ is cts on $[0,1]\times [0,1]$.

Now we have to check the two conditions

$H(s,0),H(s,1),H(0,t),H(1,t)$.

I was able to check $H(0,t)=F(0,t)=x_0$ and $H(1,t)=G(1,t)=x_2$

But I couldn't get the required answer for the other two parts $H(s,0),H(s,1)$. Can somebody please help me with this?

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    I think he may be trying to show that given $f_1,f_2\in [f]$ and $g_1,g_2\in [g]$ we have $f_1\cdot g_1\sim f_2\cdot g_2$, showing that this operation is indeed well-defined. Aka he's showing that as it's defined $f\cdot g$ is an equivalence class.2012-12-23

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I can only read Latex otherwise my eyes start to hurt but here's a comment:

Claim: If $[f]=[f']$ and $[g] = [g']$ then $[f][g] = [f'g']$.

Proof: Let $h_f$ be the homotopy $f \sim f'$ and $h_g$ be the homotopy $g \sim g'$. We now want to construct a homotopy $H: fg \sim f'g'$. It means that at $0$ you first want to traverse $f$ then $g$, similarly for $f'$ and $g'$ at $1$. If you define $H$ as follows:

$ H(x,t) = \begin{cases} h_f(2x,t) & x \in [0,\frac12] \\ h_g(2x-1,t) & x \in [\frac12 , 1] \end{cases}$

you have $H(0,t) = f (t)$ and not $f \ast g$ as desired.

I suggest that you sit down and draw a square diagram of $[0,1] \times [0,1]$ in which you draw the maps to compute what the correct definition should be.