A Concise Course in Algebraic Topology says followings:
Definition.1. Let $q$ denote positive integers. A generalized homology theory $E_{*}$ is defined to be a system of functors $E_q(X, A)$ from the homotopy category of pairs of spaces to the category of Abelian groups and natural transformations $\partial :E_q(X, A)\rightarrow E_{q-1}(A)$, where $E_q(X):=E_q(X, \varnothing)$, that satisfy the following axioms:
EXACTNESS. The following sequence is exact, where $i:A\rightarrow X$ and $j:(X, \varnothing)\rightarrow (X, A)$ are the inclusions:
$$…\rightarrow E_q(A) \xrightarrow{i_*} E_q(X) \xrightarrow{j_*} E_q(X, A) \xrightarrow{\partial} E_{q-1}(A) \rightarrow … \rightarrow E_0(X, A) \rightarrow0$$
EXCISION. If $(X; A, B)$ is an excisive triad, then the inclusion $(A, A\cap B)\rightarrow (X, B)$ induces an isomorphism $$E_*(A, A\cap B)\rightarrow E_*(X, B)$$
ADDITIVITY. If $(X, A)$ is the disjoint union of a set of pairs $(X_i, A_i)$, then the inclusions $(X_i, A_i)\rightarrow (X, A)$ induce an isomorphism $$\bigoplus_i E_*(X_i, A_i)\rightarrow E_*(X, A)$$
WEAK EQUIVALENCE. If $f:(X, A)\rightarrow (Y, B)$ is a weak equivalence, then $f_*:E_*(X, A)\rightarrow E_*(Y, B)$ is an isomorphism.
Definition.2. A reduced homology theory $\tilde{E}_*$ consists of functors $\tilde{E}_q$ from the homotopy category of nondegenerately based spaces to the category of Abelian groups that satisfy the following axioms:
EXACTNESS. If $i:A\rightarrow X$ is a cofibraion, then the following sequence is exact:
$$\tilde{E}_q(A)\rightarrow \tilde{E}_q(X) \rightarrow \tilde{E}_q(X/A)$$
SUSPENSION. There is a natural isomorphism
$$\Sigma :\tilde{E}_q(X) \rightarrow \tilde{E}_{q+1}(\Sigma X)$$
ADDITIVITY. If $X$ is the wedge of a set of nondegenerately based spaces $X_i$, then the inclusions $X_i\rightarrow X$ induce an isomorphism $$\bigoplus_i \tilde{E}_*(X_i) \rightarrow \tilde{E}_*(X)$$
WEAK EQUIVALENCE. If $f:X\rightarrow Y$ is a weak equivalence, then $f_*:E_*(X)\rightarrow E_*(Y)$ is an isomorphism.
I would like to prove the following theorem:
Theorem(p.110). A homology theory $E_*$ on pairs of spaces determines and is determined by a redeced homology theory $\tilde{E}_*$ on nondegenerately based spaces.
As the proof of the implication of "determines", given a homology theory $E_*$, then we define $\tilde{E}_q(X):=E_q(X, *)$ for any nondegenerately based space $X$ with the basepoint $*\in X$. And then we shall show the $\tilde{E}_*$ satisfies the all axioms of "Definition.2". To do this, it says
Since the basepoint is a retract of $X$, there results a direct sum decomposition
$$E_*(X)\cong \tilde{E}_*(X)\oplus E_*(*)$$
that is natural with respect to based maps. For $*\in A \subset X$, the summand $E_*(*)$ maps isomorphically under the map $E_*(A)\rightarrow E_*(X)$, and the exactness axiom implies that there is a reduced long exact sequence
$$…\rightarrow \tilde{E}_q(A) \rightarrow \tilde{E}_q(X) \rightarrow E_q(X, A) \xrightarrow{\partial} \tilde{E}_{q-1}(A) \rightarrow …$$
My question. To show the decomposition, I guess that splitting lemma will be used. Let $i:\{*\}\rightarrow X$ be the inclusion and let $r:X\rightarrow \{*\}$ be the constant map. Then the exactness axiom for $E_*$ implies the exact sequence
$$E_q(*) \xrightarrow{i_*} E_q(X) \rightarrow E_q(X, *) $$
together with $r_*\circ i_*=id$. To use splitting lemma, we must show that the exactness of $0\rightarrow E_q(*) \xrightarrow{i_*} E_q(X) $ and $E_q(X) \rightarrow E_q(X, *)\rightarrow 0 $. But I cannot prove it. How do I show it? Or, is my attempt incorrect?
For your information, it claims the following corollary associated with the theorem:
Corollary(p.111). For nondegenerately based spaces $X$, $E_*(X)$ is naturally isomorphic to $\tilde{E}_*(X)\oplus E_*(*)$.
Proof. The long exact sequence in $E_*$ of the pair $(X, *)$ is naturally split in each degree by means of the homomorphism induced by the projection $X\rightarrow \{*\}$.
Also about this, I cannot find that the two sequence $0\rightarrow E_q(*) \xrightarrow{i_*} E_q(X) $ and $E_q(X) \rightarrow E_q(X, *)\rightarrow 0 $ are exact. I thought that it could be possible that splitting lemma is proved except for the exactness at issue, and tried proving splitting lemma again, but I have been not able to do. In fact, could it be done?
Best Answer
Because $r_* \circ i_* = id$, $i_*$ is one-to-one, so the boundary map $E_q(X,*) \to E_{q-1}(*)$ is zero. That means that the long exact sequence breaks apart into short exact sequences.