Here is the problem statement I am concerned of:
For a space $A$ let $CA = A\times [0,1]/A\times \{1\}$ be the cone and if $f:X\to Y$ let $C_f = (Y\coprod CX)/\sim$ where we identify $(x,0)\sim f(x)$.
- Let $f:S^2\to S^2$ be a map of degree $3$. Compute the groups $H_i(C_f,\mathbb Z)$ for all $i\ge 0$.
- Let $g:S^2\to S^2$ be a contractible map. Compute the groups $H_i(C_g,\mathbb Z)$ for all $i\ge 0$.
Here are my questions:
a) I provided my solution for the first part as below, but I am missing one last step: what is $H_2(C_f)$? Can I also get a sanity check on my solution?
b) In the second part, I am not familiar with the terminology "contractible map". I looked it up in different sources and had no luck. Presumably, it has something to do with contractible spaces. Can anyone clarify this?
Thanks in advance!
So $C_f$ is the mapping cone of $f$. Given the mapping cone $C_f$, we divide it into $A$ and $B$, where $A$ is an open neighborhood of the tip $*$ of the cone, and $B = C_f\setminus\{*\}$. Now we use the Mayer-Vietoris sequence. Note that
- $A$ deform retracts to the tip $*$, so the homology is trivial,
- $B$ deform retracts to the bottom, i.e., $B\simeq S^2$, and
- $A\cap B$ is a cylinder with the section equivalent to the domain, so it is equivalent to $S^2$.
On each degree the sequence gives
$$
\cdots \to H_k(S^2)\to 0\oplus H_k(S^2)\to H_k(C_f)\to \cdots
$$
so whenever $k>3$ we have $H_k(C_f) = 0$. For $k =0$ obviously $H_0(C_f) =0$. For $k = 2,3$, we have
$$
0\to H_3(C_f)\to \mathbb Z\xrightarrow{\cdot 3} \mathbb Z \to H_2(C_f)\to 0
$$
where the middle map is multiplication by $3$ since this is the induced mapping by $f$, which has degree $3$. By exactness at the first $\mathbb Z$ we have $H_3(C_f)= 0$ since $H_3(C_f)\to \mathbb Z$ is injective. I am not entirely sure what $H_2(C_f)$ should be in this case though.
Best Answer
Everything you've written is correct. Your final exact sequence simplifies to $$ 0 \to \mathbb Z \xrightarrow{\cdot 3} \mathbb Z \to H_2(C_f) \to 0, $$
so by exactness $$ H_2(C_f) \cong \operatorname{coker}(\mathbb Z \xrightarrow{\cdot 3} \mathbb Z) = \mathbb Z / im(\cdot 3) = \mathbb Z/3\mathbb Z, $$
where the middle equality is just the definition of cokernel.
For $H_1$, we have $$ 0 \to H_1(C_f) \xrightarrow f \mathbb Z \xrightarrow g \mathbb Z \xrightarrow h \mathbb Z \to 0 $$
with $h$ surjective hence $\pm id$, so by exactness at the middle $\mathbb Z$ we have $\operatorname{im} g = \ker h = 0$ so $g$ is the $0$ map. Thus $\operatorname{im} f = \ker g = \mathbb Z$, making $f$ an isomorphism.