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.
Best Answer
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.