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[SOLVED] homotophy

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $\gamma$ be a closed curve in U with initial point $z_0$, and let $\gamma^-$ denote its reverse curve.
Prove that $\gamma\gamma^-$ is null homotopic in U.

So I need to show that $\int_{\gamma\gamma^-}f(z)dz = 0$. How can I show this though? Not all closed curves are necessarily zero.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Let $\gamma$ be a closed curve in U with initial point $z_0$, and let $\gamma^-$ denote its reverse curve.
Prove that $\gamma\gamma^-$ is null homotopic in U.

So I need to show that $\int_{\gamma\gamma^-}f(z)dz = 0$. How can I show this though? Not all closed curves are necessarily zero.
Rather than using the property $\int_{\gamma\gamma^-}f(z)dz = 0$, I think it would be easier to construct an explicit homotopy from $\gamma\gamma^-$ to the constant path at $z_0.$

Given $\gamma:[0,1]\to U$, with $\gamma^-(t) = \gamma(1-t)$, define $H_s$ for $s\in[0,1]$ by $$H_s(t) = \begin{cases}\gamma^-(2st)&(t\leqslant1/2),\\ \gamma(2st)&(t\geqslant1/2).\end{cases}$$ Then $H_1=\gamma\gamma^-$ and $H_0$ is the constant path at $z_0.$

Of course, you need to check that, for each $s$, $H_s$ is a path in $U$, and that the map $s\mapsto H_s$ is indeed a homotopy.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Rather than using the property $\int_{\gamma\gamma^-}f(z)dz = 0$, I think it would be easier to construct an explicit homotopy from $\gamma\gamma^-$ to the constant path at $z_0.$

Given $\gamma:[0,1]\to U$, with $\gamma^-(t) = \gamma(1-t)$, define $H_s$ for $s\in[0,1]$ by $$H_s(t) = \begin{cases}\gamma^-(2st)&(t\leqslant1/2),\\ \gamma(2st)&(t\geqslant1/2).\end{cases}$$ Then $H_1=\gamma\gamma^-$ and $H_0$ is the constant path at $z_0.$

Of course, you need to check that, for each $s$, $H_s$ is a path in $U$, and that the map $s\mapsto H_s$ is indeed a homotopy.
How did you come up with this piece wise defined map?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
How did you come up with this piece wise defined map?
Would this map work as well?

$$
F(s,t)=\begin{cases}\gamma(2s,t), & 0\leq s\leq 1/2\\
\gamma^-(2s-1,t), & 1/2\leq s\leq 1
\end{cases}
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $[\gamma]$ be the homotopy class in $\pi_1(U,z_0)$.
Let
$$
\gamma\gamma^- = \begin{cases}
\gamma(2t), & 0\leq t\leq\frac{1}{2}\\
\gamma(2 - 2t), & \frac{1}{2}\leq t\leq 1
\end{cases}.
$$
Then the path $\gamma\gamma^-$ follows $\gamma$ and then $\gamma^-$ back to $z_0$.
Define the homotopy
$$
K(t,s) = \begin{cases}
\gamma(2ts), & 0\leq t\leq\frac{1}{2}\\
\gamma(2s - 2ts), & \frac{1}{2}\leq t\leq 1.
\end{cases}
$$
Then $K(0,2) = K(1,s) = z_0$, $K(t,0) = c$ (the constant path), and
$$
K(t,1) = \begin{cases}
\gamma(2ts), & 0\leq t\leq\frac{1}{2}\\
\gamma(2 - 2t), & \frac{1}{2}\leq t\leq 1
\end{cases}
= \gamma\gamma^-.
$$
We can observe that $K$ is continuous.
Now $[\gamma][\gamma^-] = [\gamma\gamma^-] = [c]$.
Thus the product of the classes of $\gamma$ and $\gamma^-$ is the constant path, i.e. null homotopic.

How is this?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Looks good. :)