# Problem of the Week #367 - June 28, 2022

• MHB
• Euge

#### Euge

Gold Member
MHB
POTW Director
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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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.

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.

Last edited by a moderator:
Greg Bernhardt
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!

Greg Bernhardt
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}##.