I'm reading Tom Dieck's "Algebraic Topology" but get stuck on proving the claim (in l.5 of p. 247):
Claim A. If $X$ and $A$ are contractible, then $h_n(X,A)=0$.
where $h_{n}\colon\mathrm{Top}(2)\to\mathrm{Ab}$ is a homology theory satisfying
- Homotopy invariance with respect to homotopy of $\mathrm{Top}(2)$.
- Long exactsequence: For $(X,A)\in\mathrm{Top}(2)$, we have an exact sequence
$$
\cdots
\to h_{n+1}(X,A)
\overset{\delta}{\to} h_{n}(A, \emptyset)
\to h_n(X,\emptyset)
\to h_n(X,A)
\to \cdots
$$
where undecorated morphisms are induced by inclusions and the connecting morphism $\delta$ is natural with respect to $(X,A)$. - Excision.
I can prove the following modified claim:
Claim B. If $(X,A)$ is contractible in $\mathrm{Top}(2)$ in the sense that there is a homotopy
$$
F\colon (X\times I, A\times I)\to (X\times I, A\times I)
$$
from $\mathrm{id}_{(X,A)}$ to the constant map $(X,A)\to (X,A); x\mapsto a\in A$,
then $h_n(X,A)=0$.
Indeed, by the homotopy invariance, we have $h_n(X,A)\cong h_n(\{a\},\{a\})$ and then the long exact sequence with $(\{a\},\{a\})\in\mathrm{Top}(2)$ gives $h_n(\{a\},\{a\})=0$.
The difference of Claim A and Claim B is whether or not the contractibility is compatible between $X$ and $A$, i.e. there is a simultaneous contraction $F\colon X\times I\to X$ of $X$ and $A$. My question is thus
Q. Is it possible to prove Claim A without such compatibility?
If the answer is not, then I would like to know some simple counter-example preferably in the relative singular homology theory.
Best Answer
If you read the paragraph in Dieck right before where you're having issues, note that it says that if $f \colon (X, A) \to (Y, B)$ is a map such that the components $f \colon X \to Y$ and $f \colon A \to B$ induce isomorphisms on homology, then the induced maps $f_{*} \colon h_{n}(X, A) \to h_{n}(Y, B)$ are isomorphisms.
Take $f \colon (X, A) \to (\text{pt}, \text{pt})$ to be the map of pairs whose components are the constant maps $X \to \text{pt}$, $A \to \text{pt}$. Since $X$ is contractible, it is homotopy equivalent to a point, and $f \colon X \to \text{pt}$ must be a homotopy equivalence (because there are no other maps!). Similarly, $f|_{A} \colon A \to \text{pt}$ must be a homotopy equivalence.
We know that homotopy equivalences induce isomorphisms on homology. So, putting the two paragraphs together, we have $$h_{n}(X, A) \cong h_{n}(\{\text{pt}\}, \{\text{pt}\}) = 0.$$ Note that we don't actually have to worry about any explicit homotopies.