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Show that composition of paths satisfies the following cancellation property: if $f_0 \cdot g_0 \simeq f_1 \cdot g_1 $ and $g_0 \simeq g_1$, then $f_0 \simeq f_1$.

So I have two homotopies.

So say $g_0,g_1: X \rightarrow Y$ and $f_0,f_1:Y \rightarrow Z$.

Then we know that $G:X \times I \rightarrow Z$ s.t. $G(x,0)=f_0 \cdot g_0(x)$ and $G(x,1)=f_1 \cdot g_1 (x)$. Also, $H:X \times I \rightarrow Y$ s.t. $H(x,0)=g_0(x)$ and $H(x,1)=g_1(x)$.

I was wondering how do you construct the homotopy for f? or is there a simplier way.

I would think you construct a homotopy, but can't see how to.

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Wait: by "composition of paths" do you mean concatenation? By definition a path a continuous function $\gamma:[0,1]\rightarrow X$, so what do you mean by "$g_0,g_1:X\rightarrow Y$ and $f_0,f_1:Y\rightarrow Z$"? –  you Apr 5 '12 at 23:16
    
All you need to show is that if $h_0 \simeq h_1$ and $k_0 \simeq k_1$ then $h_0 \cdot k_0 \simeq h_1 \cdot k_1.$ Thus, now take, $k_i=f_i \circ g_i$ and $h_i=g^{-1}_i.$ –  Ehsan M. Kermani Apr 5 '12 at 23:19
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@you is that concatenation? hmm that probably where I've been going wrong. Thought it might be concatenation,hmm. –  simplicity Apr 5 '12 at 23:21
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3 Answers 3

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It helps if you first show that inverses of homotopic paths are homotopic, which is relatively easy to do from the definitions. That is, show that if $f_1 \simeq f_2$, then $\bar{f_1} \simeq \bar{f_2}$. Once you have that, start with $f_0 \simeq f_0 \cdot (g_0 \cdot \bar{g_0}) \simeq (f_0 \cdot g_0) \cdot \bar{g_0}$, and notice that there's a nice copy of $f_0 \cdot g_0$ in there.

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Translate this to the fundamental groupoid having the homotopy classes of the paths as its elements.

You are asked to show that $\left[f_{0}\right]\left[g_{0}\right]=\left[f_{1}\right]\left[g_{1}\right]\wedge\left[g_{0}\right]=\left[g_{1}\right]$ implies that $\left[f_{0}\right]=\left[f_{1}\right]$, or shorter: $\left[f_{0}\right]\left[g\right]=\left[f_{1}\right]\left[g\right]\Rightarrow\left[f_{0}\right]=\left[f_{1}\right]$.

Then realize that $\left[g\right]$ has an inverse (as any element in a groupoid). That will do. In my view this is indeed a simpler way.

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Since $f \cdot g(s) = f(2s){\kern 1pt} {\kern 1pt} {\kern 1pt} for{\kern 1pt} 0 \le s \le \frac{1}{2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ;g(2s - 1){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} for{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{2} \le s \le 1 $ and $G$ is a homotopy between $f_0 \cdot g_0 $and $f_1 \cdot g_1$.we could find a homotopy $F$ between $f_0$ and $f_1$ by $F(s)=G(1/2 s)$

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I didn't use the homotopy between $g_0$ and $g_1$,maybe I'm wrong. –  Jiangnan Yu Apr 6 '12 at 7:21
    
Your map $F$ isn't defined at $0$. –  Holdsworth88 Aug 22 '12 at 7:10
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