# How can a topological space be homotopic to a point?

I have a problem in algebraic topology:

We defined Homotopy between spaces like this:

Two top. spaces are called homotopic equivalent, if there are two continuous maps $$f:X\to Y$$ and $$g:Y\to X$$ s.t. $$(f\circ g)(x)\simeq 1_Y$$ and $$(g\circ f)(x)\simeq 1_X$$.

So my question is: How can a top. space $$X$$ be homotopic to a point (i.e. contractible), if it has more than one point?

because if we take that one point as $$Y$$ in the definition of homotopic, we can't find $$f$$ and $$g$$ like that, because $$g$$ can have only one value, because it's domain is one point.

• That's not the right definition. You have to require that the compositions of $f$ and $g$ be homotopic to the identity (and also that $f,g$ be continuous). May 29 '20 at 5:58
• Your definition of homotopic space has has two mistakes. First of all your condition is equivalent ti $X$ and $Y$ having the same cardinality. Secondly, even if you insist on $f$ and $g$ being continuous you get homemomorphic spaces. The definition has nothing to do with homotopy. May 29 '20 at 5:59
• I didn't know how to make that little circle for composition so i wrote the composition like this f(g(x)). It's the same thing right? And you are right i should have mention it's continuous (functions are assumed to be continuous unless otherwise stated in this course) May 29 '20 at 6:08
• Oh sorry now i get it, the composition doesn't need to be equal to the identity only homotopic. Sorry i copied wrong from the lecture probably. Should i delete the question? May 29 '20 at 6:14
• Also, $f\circ g$ is not the same as $f(g(x))$. The first one is a function from $Y$ to $Y$, while the second is a point of $Y$. Btw, the circle is \circ. May 29 '20 at 6:52

Take simple example: $$X=\{0\}$$ and $$Y=[0,1]$$ the standard interval with the Euclidean topology.

Now define $$f:X\to Y$$, $$f(0)=0$$ and $$g:Y\to X$$ by (not much choice here) $$g(x)=0$$.

So we now consider the composition $$g\circ f$$ which is a function $$X\to X$$ given by $$g\circ f(0)=0$$. This is not only homotopic to the identity but it is the identity itself.

On the other hand consider $$f\circ g:Y\to Y$$. This time $$f\circ g(x)=0$$ is a constant function. So we need to show that it is homotopic to the identity. For that consider

$$H:I\times Y\to Y$$ $$H(t, x)=tx$$

Obviously $$H$$ is continuous, $$H(1,x)=x$$ and $$H(0,x)=0$$. And so $$H$$ is a homotopy from $$f\circ g$$ to the identity.

So as you can see, unlike homeomorphisms, homotopy equivalences do not have to preserve cardinality. Moreover homotopy equivalences don't have to be injective or surjective. In fact if $$X,Y$$ are contractible then any continuous function $$X\to Y$$ is a homotopy equivalence.