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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?

  • 0
    You're trying to "prove that $[f]*[g]=[f*g]$"? Isn't that the definition of $[f]*[g]$?2012-12-23
  • 3
    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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