Problem of the Week #367 - June 28, 2022

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Here is this week's problem!

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Let $X$ be a topological space. Show that $X$ is homotopy equivalent to the cylinder $X \times [0,1]$.
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[Office_Shredder]

I would have thought a cylinder would be $X\times S^1$. I'm going to assume it's the space stated since otherwise the statement isn't true I think.

Spoiler

We need to construct maps $f: X\to X\times [0,1]$ and $g: X\times [0,1] \to X$ such that $g\circ f: X \to X$ and $f\circ g: X\times[0,1]\to X\times [0,1]$ are each homotopy equivalent to the identity maps on these spaces.

Let $f(x)=(x,0)$ and $g(x,t)=x$. Then $g(f(x))=x$ is already the identity map. So we need to prove $f(g((x,t)))=(x,0)$ is homotopy equivalent to the identity map on $X\times [0,1]$.
Let's call $f\circ g=h$ and $X\times [0,1]=Y$. To show $h$ is homotopy equivalent to the identity map $id$, we need to construct $H(y,s) :Y\times [0,1] \to Y\times [0,1]$ such that $H$ is continuous $H(y,0)=h(y)$ and $H(y,1)=id(y)$.
I'm going to drop the $y$/$Y$ notation now.


Let $H( (x,t),s)=(x,ts)$ then $H$ is continuous, $H((x,t),1)=(x,t)$ is the identity map, and $H((x,t),0)=(x,0)$ is $f\circ g$. Hence these two maps are homotopy equivalent, completing the proof that $X$ and $X\times [0,1]$ are homotopy equivalent.

 

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There was a typo in the Problem statement: the term “cylinder” should have been replaced with “product.” In any case, your solution is correct. Thanks for participating!

 

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Just a side note: The space $X\times \mathbb{S}^1$ is not generally homotopy equivalent to $X$. For if $X = \mathbb{S}^1$, then the fundamental group of $X\times \mathbb{S}^1$ is $\mathbb{Z}\times \mathbb{Z}$ whereas the fundamental group of $X$ is $\mathbb{Z}$.