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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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    @you is that concatenation? hmm that probably where I've been going wrong. Thought it might be concatenation,hmm.2012-04-05

4 Answers 4

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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.

3

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.

0

If $X_0$ is the path-component of a space $X$ containing the basepoint $x_0$, show that the inclusion $X_0֓\subset X$ induces an isomorphism $\pi_1(X_0,x_0)\to\pi_1(X,x_0)$.

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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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    Your map $F$ isn't defined at $0$.2012-08-22