Map homotopic to constant induces the zero homomorphism

Question

Question: Show that if $f:X\to Y$ is homotopic to a constant, then $f_*$ is the zero homomorphism in reduced homology. Conclude that if $X$ is contractible, then $X$ is acyclic in singular homology.

My thoughts: So this seems pretty intuitive. Since $f$ is homotopic to a constant, this map can be seen as mapping all of $X$ into a single point of $Y$, so it makes sense that the induced homomorphism is the zero homomorphism; i'm having trouble making this rigorous.

Furthermore, as to conclude that $X$ beign contractible implies that $X$ is acyclic in singular homology: would I take a map $g: X\to X$ that is a contraction of $X$, and then, since homotopy equivalence induces isomorphic homology groups, it would follow that $X$ is acyclic?

Insights and clarifications are greatly appreciated!!

Answer 1

It is a general fact that homotopic maps induce the same maps on homology. Basically the idea is that homotopic maps induce chain homotopic maps, which in turn induce the same map on homology. If you know that maps that are actually constant induce the $0$ map on homology, then your claim follows.

To prove this general fact, we have to show two things. That homotopic maps induce chain homotopic maps, and that chain homotopic maps induce the same map on homology.

The first can be found here.

The second can be found here.

Probably these are standard facts that you are expected to know, so it might be worthwhile to digest them.

Answer 2

You can make more of your statement "...this map can be seen as mapping all of $X$ into a single point of $Y$." In particular, you can use this idea to factorise the map $f$ up to homotopy as

$f\simeq j\circ q:X\xrightarrow{q}\ast\xrightarrow{j} Y$

where the first map $q$ collapses all of $X$ to a point, and the second map $j$ is the inclusion of some point of $Y$ lying in the path component of the image of $f$.

By the homotopy invariance and functorality of reduced homology you have for the induced map $f_*$ that

$f_*=(j\circ q)_*=j_*\circ q_*:\tilde H_*(X)\rightarrow \tilde H_*(\ast)\rightarrow \tilde H_*(Y).$

Hence the homomorphism $f_*$ factors through the reduced homology of the 1-point space, $\tilde H_*(\ast)=0$, so in particular $f_*=0$.

Now if $X$ is contractible, then the identity map on $X$ factors

$id_X\simeq j\circ q:X\rightarrow \ast\rightarrow X$

Then you have by homotopy invariance and functorality that

$id_{\tilde H(X)}=(id_{X})_*=(j\circ q)_*=0$

and the only way that the identity $id_{\tilde H(X)}$ of the singular chain complex $\tilde H_*(X)$ can be the zero homomorphism is if $\tilde H_*(X)$ is itself the zero group. Hence $X$ is acyclic.