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- #1

I am pretty lost on this problem.

- Thread starter dwsmith
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- Thread starter
- #1

I am pretty lost on this problem.

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- #2

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- #3

$$

\Gamma_0=\begin{cases}\gamma_0(t,2s), & 0\leq s\leq\frac{1}{2}\\

\delta_0(t,2s-1), & \frac{1}{2}\leq s\leq 1

\end{cases}

$$

$$

\Gamma_1=\begin{cases}\gamma_1(t,2s), & 0\leq s\leq\frac{1}{2}\\

\delta_1(t,2s-1), & \frac{1}{2}\leq s\leq 1

\end{cases}

$$

So the product curve should be?

$$

\gamma_0\gamma_1=\begin{cases}\gamma_0(2t), & 0\leq t\leq\frac{1}{2}\\

\underbrace{\gamma_1(2t)}_{2t \ \text{as well?}}, & \frac{1}{2}\leq t\leq 1

\end{cases}

$$

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- #4

I believe this should work and make the product curves homotopic to a point.

Given that $\Gamma_0$ and $\Gamma_1$ are homotopies let

$$

\gamma_0\gamma_1(t,s) =

\begin{cases}\gamma_0(2ts), & 0\leq t\leq\frac{1}{2}\\

\gamma_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1

\end{cases}

\quad\text{and}\quad

\delta_0\delta_1(t,s) =

\begin{cases}\delta_0(2ts), & 0\leq t\leq\frac{1}{2}\\

\delta_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1

\end{cases}.

$$

Since the direction of neither $\gamma_0$ and $\gamma_1$ nor $\delta_0$ and $\delta_1$ were specified, we can have $\gamma_1$ and $\delta_1$ run in the reverse direction as $\gamma_0$ and $\delta_1$ but they aren't specifically the same path.

Then $\gamma_0\gamma_1(0,s) = \gamma_0\gamma_1(1,s) = z_0$, $\gamma_0\gamma_1(t,0) = \gamma_0$, and $\gamma_0\gamma_1(t,1) = \gamma_1$.

Also, $\delta_0\delta_1(0,s) = \delta_0\delta_1(1,s) = z_0$, $\delta_0\delta_1(t,0) = \delta_0$, and $\delta_0\delta_1(t,1) = \delta_1$.

If $U$ is simply connected, then the product curves would be homotopic to a point.

If $U$ isn't simply connected such that two curves wrapped around a doughnut, then the product curves would be path homotopic.

Given that $\Gamma_0$ and $\Gamma_1$ are homotopies let

$$

\gamma_0\gamma_1(t,s) =

\begin{cases}\gamma_0(2ts), & 0\leq t\leq\frac{1}{2}\\

\gamma_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1

\end{cases}

\quad\text{and}\quad

\delta_0\delta_1(t,s) =

\begin{cases}\delta_0(2ts), & 0\leq t\leq\frac{1}{2}\\

\delta_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1

\end{cases}.

$$

Since the direction of neither $\gamma_0$ and $\gamma_1$ nor $\delta_0$ and $\delta_1$ were specified, we can have $\gamma_1$ and $\delta_1$ run in the reverse direction as $\gamma_0$ and $\delta_1$ but they aren't specifically the same path.

Then $\gamma_0\gamma_1(0,s) = \gamma_0\gamma_1(1,s) = z_0$, $\gamma_0\gamma_1(t,0) = \gamma_0$, and $\gamma_0\gamma_1(t,1) = \gamma_1$.

Also, $\delta_0\delta_1(0,s) = \delta_0\delta_1(1,s) = z_0$, $\delta_0\delta_1(t,0) = \delta_0$, and $\delta_0\delta_1(t,1) = \delta_1$.

If $U$ is simply connected, then the product curves would be homotopic to a point.

If $U$ isn't simply connected such that two curves wrapped around a doughnut, then the product curves would be path homotopic.

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I think you just want $\Gamma(t,s):=\begin{cases}\Gamma_0(t,2s)&&s\le \frac12\\\Gamma_1(t,2s-1)&&s\ge \frac12\end{cases}$, where $t$ is time (the homotopy parameter) and $s$ is going around the curve. In other words, at time $t$ what you have is the concatenation of something between $\gamma_0$ and $\delta_0$ with something between $\gamma_1$ and $\delta_1$. Graphically, if each homotopy is a function on the square $I\times I$, then what we did was shrink them horizontally so they are functions on $[0,\frac12]\times I$ and $[\frac12,1]\times I$, then we glued them together along $t=\frac12$, where they have the same values (because all the loops have the same base point). So to explain why the result is a homotopy, you just need to invoke the gluing lemma.

PS: My first post, and I'm already in love with this forum because I can use dollar-signs for TeX! Sweet freedom!

PS: My first post, and I'm already in love with this forum because I can use dollar-signs for TeX! Sweet freedom!

Last edited:

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- #6

Even by using the pasting lemma, how does that show the product curves are homotopic?I think you just want $\Gamma(t,s):=\begin{cases}\Gamma_0(t,2s)&&s\le \frac12\\\Gamma_1(t,2s-1)&&s\ge \frac12\end{cases}$, where $t$ is time (the homotopy parameter) and $s$ is going around the curve. In other words, at time $t$ what you have is the concatenation of something between $\gamma_0$ and $\delta_0$ with something between $\gamma_1$ and $\delta_1$. Graphically, if each homotopy is a function on the square $I\times I$, then what we did was shrink them horizontally so they are functions on $[0,\frac12]\times I$ and $[\frac12,1]\times I$, then we glued them together along $t=\frac12$, where they have the same values (because all the loops have the same base point). So to explain why the result is a homotopy, you just need to invoke the gluing lemma.

PS: My first post, and I'm already in love with this forum because I can use dollar-signs for TeX! Sweet freedom!

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- #8

But we want $\gamma_0\gamma_1$ to be the homotopic. We are given that those are homotopic already.By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\delta_0$ and $\Gamma(1,\cdot)=\gamma_1\delta_1$. This, together with continuity, is the definition of path homotopy.

By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\gamma_1$ and $\Gamma(1,\cdot)=\delta_0\delta_1$. This, together with continuity, is the definition of path homotopy.

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- #10

So this part is true via the pasting lemma?

By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\gamma_1$ and $\Gamma(1,\cdot)=\delta_0\delta_1$. This, together with continuity, is the definition of path homotopy.

So $\Gamma$ is the homotopy we're after,

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- #12

Shouldn't that be the other way around. If t = 0, shouldn't we be on $z_0$ and when s = 0, we should be on the first curve?

So $\Gamma$ is the homotopy we're after,ifit's continuous. The pasting lemma gives you a sufficient condition for such a "piecewise-defined" function to be continuous, and $\Gamma$ satisfies the condition.

We just concatenated the homotopies, by concatenating each path in one with the path in the other corresponding to the same $t$ value.