# Can a Topological Space Be Homotopy Equivalent to a Cylinder?

• MHB
• Euge
In summary, a homotopy equivalence is a continuous function between two topological spaces that can be continuously deformed into each other. A topological space is a set of points with specific properties that allow for the study of continuous functions. A cylinder and a sphere cannot be homotopy equivalent because they have different fundamental groups. Two spaces are homotopy equivalent if they have the same homotopy type, which is determined by topological invariants such as the fundamental group. To prove that a space is homotopy equivalent to a cylinder, one must show that the fundamental group, homology groups, and homotopy groups of the two spaces are the same.
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}##.

## 1. Can a topological space be homotopy equivalent to a cylinder?

Yes, it is possible for a topological space to be homotopy equivalent to a cylinder. This means that the two spaces have the same homotopy type, which is a measure of their connectivity and deformation properties.

## 2. What does it mean for two spaces to be homotopy equivalent?

Two spaces are homotopy equivalent if there exists a continuous map between them that can be continuously deformed into a homotopy equivalence. This means that the two spaces have the same fundamental group, homology groups, and other homotopy invariants.

## 3. How can a topological space be homotopy equivalent to a cylinder?

A topological space can be homotopy equivalent to a cylinder if it has the same homotopy type as a cylinder. This can be achieved by continuously deforming the space into a shape that is topologically equivalent to a cylinder.

## 4. What are the implications of two spaces being homotopy equivalent to a cylinder?

If two spaces are homotopy equivalent to a cylinder, it means that they have the same homotopy invariants and can be continuously deformed into each other. This can have implications in various areas of mathematics, such as algebraic topology and differential geometry.

## 5. Can a topological space be homotopy equivalent to a cylinder if it is not a cylinder itself?

Yes, a topological space can be homotopy equivalent to a cylinder even if it is not a cylinder itself. This is because homotopy equivalence is a measure of the spaces' connectivity and deformation properties, not their specific shape or structure.

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