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.

  • 4
    $\begingroup$ 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). $\endgroup$
    – peek-a-boo
    May 29 '20 at 5:58
  • 4
    $\begingroup$ 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. $\endgroup$ May 29 '20 at 5:59
  • $\begingroup$ 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) $\endgroup$
    – Ton910
    May 29 '20 at 6:08
  • $\begingroup$ 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? $\endgroup$
    – Ton910
    May 29 '20 at 6:14
  • $\begingroup$ 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. $\endgroup$ 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.


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