$E_*(X)\cong \tilde{E}_*(X)\oplus E_*(*)$ from Eilenberg–Steenrod axioms 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?
 A: 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.
A: Since it is a good opportunity, I will write the proof of the above theorem. I hope this helps. And please tell me if my proof is mistaken.

Fact(p.45). Any continuous map is a composite of a homotopy equivalence and a cofibration.


Lemma(p.108). For any cofibration $i:A\rightarrow X$, the quotient map $q:(X, A)\rightarrow (X/A, *)$ induces an isomorphism $E_*(X,A)\rightarrow E_*(X/A, *)$.


Theorem(p.110). A homology theory $E_*$ on pairs of spaces determines and is determined by a reduced homology theory $\tilde{E}_*$ on nondegenerately based spaces.

Proof. We choose a homotopy inverse $\psi^{-1}:X/A\rightarrow Ci$ and consider the composite $X/A\xrightarrow{\psi^{-1}}Ci\xrightarrow{\pi}\Sigma A$ to be a topological map $\partial:X/A\rightarrow \Sigma A$, where $i:A\rightarrow X$ is the inclusion and $\pi$ is the quotient map.
Show that $E_*$ determines $\tilde{E}_*$: Given a homology theory $E_*$, we define $\tilde{E}_q(X):=E_q(X,*)$ for any based space $X$.
EXACTNESS) Appling the pair $(X,*)$ to the exactness axiom, we have the long exact sequence
$$...\rightarrow E_q(*) \xrightarrow{i_*} E_q(X) \rightarrow \tilde{E}_q(X) \xrightarrow{\partial} E_{q-1}(*) \rightarrow ... \rightarrow \tilde{E}_0(X) \rightarrow0$$
Let $r:X\rightarrow \{*\}$ be the constant map. Since $r_*\circ i_*=id$, $i_*$ is injective. So $im\partial =keri_* =0$ and thus it gives the split exact sequence
$$0\rightarrow E_q(*) \xrightarrow{i_*} E_q(X) \rightarrow \tilde{E}_q(X) \xrightarrow{\partial} 0$$
together with $r_*\circ i_*=id$. It follows from splitting lemma that $E_*(X)\cong \tilde{E}_*(X)\oplus E_*(*)$. 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
$$...\rightarrow \tilde{E}_q(A) \rightarrow \tilde{E}_q(X) \rightarrow E_q(X, A) \xrightarrow{\partial} \tilde{E}_{q-1}(A) \rightarrow ...$$
If $i:A\rightarrow X$ is a cofibration, the previous lemma deduces the desired exact sequence.
SUSPENSION) Observe that $\tilde{E}_q(*)=0$ for all $q\neq 0$ because of the long exact sequence
$${E}_q(*) \xrightarrow{id} {E}_q(*) \rightarrow \tilde{E}_q(*) \xrightarrow{\partial} {E}_{q-1}(*) \xrightarrow{id} {E}_{q-1}(*)$$
Since $CX$ is contractible, we have the reduced exact sequence for the pair $(CX, X)$
$$0\rightarrow \tilde{E}_q(CX/X) \xrightarrow{\partial} \tilde{E}_{q-1}(X) \rightarrow 0$$
and an isomorphism $\tilde{E}_q(\Sigma X)\xrightarrow{{\partial_*}^{-1}} \tilde{E}_q(CX/X)\xrightarrow{\partial} \tilde{E}_{q-1}(X)$ since the topological map induces an isomorphism by the lemma. We define $\Sigma:=\partial_* \circ \partial^{-1}$ that is natural since the connecting homomorphism is natural transformations and the topological map has the naturality of quotient maps.
ADDITIVITY) Given the wedge $X$, we find $X=\coprod_i X_i/ \coprod_i *_i$. The additivity axiom and the lemma imply
$$\bigoplus_i E_*(X_i,*_i)\cong E_*(\coprod_i X_i, \coprod_i *_i)\cong E_*(X,*)$$
WEAK EQUIVALENCE) This is immediate by the weak equivalence axiom for $E_*$.
Show that $E_*$ is determined by $\tilde{E}_*$: Assume given a reduced homology theory $\tilde{E}_*$. By the previous fact, we may redefine functors $\tilde{E}_q$ on the homotopy category of based spaces and identify any inclusion $i:A\rightarrow X$ with a cofibration. We define $E_q(X):=\tilde{E}_q(X_+)$ and $E_q(A,X):=\tilde{E}_q(C(i_+))$, where $C(i_+)$ is the unreduced cofiber of the based inclusion $i_+:A_+\rightarrow X_+$.
EXACTNESS) Since $X_+/A_+=X/A$ is homotopic equivalent to $C(i_+)$, $\tilde{E}_q(X_+/A_+)$ is isomorphic to $E_q(X,A)$. We define the connection homomorphism $\partial_q:E_q(X,A)\rightarrow E_{q-1}(A)$ to be the composite $\tilde{E}_q(X_+/A_+)\xrightarrow{\partial_*} \tilde{E}_q(\Sigma A_+)\xrightarrow{\Sigma^{-1}}\tilde{E}_{q-1}(A_+)$, which is natural. The exactness axiom implies that the each top is exact in the following diagrams:
$\require{AMScd}$
\begin{CD}
\tilde{E}_q(X_+) @>>> \tilde{E}_q(C(i_+))  @>\pi_*>> \tilde{E}_q(\Sigma A_+) \\
@|               @A (\psi^{-1})_* A \cong A       @V \Sigma^{-1}V \cong V \\
\tilde{E}_q(X_+) @>>> \tilde{E}_q(X_+/A_+) @>\partial_q>> \tilde{E}_{q-1}(A_+)
\end{CD}
\begin{CD}
\tilde{E}_q(C(i_+)) @>i_*>> \tilde{E}_q(Ci)  @>\pi_*>>  \tilde{E}_q(\Sigma X_+) \\
@|               @V \psi_* V \cong V       @| \\
\tilde{E}_q(C(i_+)) @> \pi_* >> \tilde{E}_q(\Sigma A_+) @>(\Sigma i_+)_*>> \tilde{E}_q(\Sigma X_+) \\
@A(\psi^{-1})_*A\cong A           @V \Sigma^{-1} V \cong V       @V \Sigma^{-1}V \cong V \\
\tilde{E}_q(X_+/A_+) @>\partial_q>> \tilde{E}_{q-1}(A_+) @>(i_+)_*>> \tilde{E}_{q-1}(X_+)
\end{CD}
Here the second diagram is obtained by the homotopy diagram on page 60. By the commutatively, the each bottom is exact, as desired.
ADDITIVTY) Given the disjoint union $(X,A)$, then $X/A \cong \bigvee_i X_i/A_i$ and thus the additivity axiom and the lemma directly say the desired isomorphism.
WEAK EQUIVALENCE) If $f:(X,A)\rightarrow (Y,B)$ is a weak equivalence, so is the map $f:C(i_+)\rightarrow C(j_+)$, where $i:A\rightarrow X$ and $j:B\rightarrow Y$ are the inclusions.
EXCISION) By the weak equivalence axiom, we may assume that $(X;A,B)$ is a CW triad. Under our hypothesis, it is obvious since $A/A\cap B$ is isomorphic to $X/B$ as CW complexes. qed.
