I have to think about it and don't have the time, but I would say the following:
- I would start with $g =1$. That's the standard trick. :-)
- In this case, I would remember that your morphism $\Phi$ is what I called in my answer $(i_*, j_*)$, the one induced by the inclusions $A \leftarrow A \cap B \rightarrow B$.
- Then, we can take the generators for the $H_1$ of $A \cap B = M_1 = \mathbb{T}^2$ to be two $S^1$: one "meridian" and one "parallel". Take the equatorial inner one for the later, for instance.
- What happens with these generators inside of $A = B= R$? (That is, once we apply $i_*$ and $j_*$ to them?) Well, the first one, the meridian, goes to zero and the second one, the parallel, survives as himself. Doesn't them?
So, you've got your morphism $\Phi = (i_*, j_*)$ and you can pursue your computations, I think.
EDIT. You've probably already guessed it, but for $g=2$ , you have four generators: two meridians and two parallels too. Meridians go to zero through $(i_*, j_*)$. Parallels remain the same. And so on: in general, you'll have $2g$ generators...
Is there any difference between the reduced and unreduced suspension? This is related with this arguing here: https://mathoverflow.net/questions/107430/does-the-reduced-mapping-cylinder-have-the-same-homotopy-type-of-unreduced-mappin
The reduced cone has the same homotopy type of the unreduced one, when we are talking about well based spaces. The problem is if the inclusion $X\to CX $, in which $CX$ is the reduced cone, is a unbased cofibration. (I don't think it is in general: only if it is a well pointed space).
Assuming it is well based space, you don't have to worry if it is the reduced suspension or the unreduced one.
Back to the problem:
We are only considering unreduced constructions, as I guess you want.
This is consequence of the excision and the exact sequence.
I guess you are talking about Eilenberg-Steenrod axioms for unreduced homology, since the Eilenberg-Steenrod axioms for the reduced homology assumes the suspension isomorphism as an axiom.
$(\sum X , CX , CX) $ is a excisive triad. By the axiom, you get that
$(CX,X)\to (\sum X, CX)$ induces isomorphisms between the homology groups.
Well, now, using the exact sequence of the pair $(CX, X) $, we can prove that
$ H_q(X, \ast )$ is isomorphic to $H_{q+1}(CX,X) $.
TO prove this last statement, first of all, you have to notice that $ H_q(X)\cong H_q(X, \ast)\oplus H_q(\ast ) $. And, then, notice that that we get a new exact sequence from this congruence. This new exact sequence is
$\cdots \rightarrow H_q(A, \ast )\to H_ q(X, \ast )\to H_q(X,A)\to H_{q-1}(X, \ast)\rightarrow \cdots $.
In our case,
$\cdots \rightarrow H_q(X, \ast )\to H_ q(CX, \ast )\to H_q(CX,X)\to H_{q-1}(X, \ast)\rightarrow \cdots $.
Since $H_ q(CX, \ast )$ is clearly trivial, we conclude that
$H_q(CX,X)\to H_{q-1}(X, \ast) $
is an isomorphism.
So we can conclude that there is an isomorphism $H_q(X, \ast)\to H_ {q+1}(\sum X, CX) $.
Since $ (\sum X, CX)\equiv (\sum X, \ast) $, the required statement was proven.
Best Answer
Consider a pair $(X \lor Y, X)$. Let $Z = X - U$. By excision axiom, inclusion $(X \lor Y - Z, X - Z) \subset (X \lor Y, X)$ induces isomorphism in homology. As we have strong deformation retraction of $U$ on $x_0$, and since $(X \lor Y - Z, X - Z) = (U \lor Y, U)$, we get a deformation retraction of a pair $(U \lor Y, U) \to (Y, x_0)$, and the homotopy axiom says that this induces isomorphism in homology. The inverse is given by the map given by inclusion. Overall, we have that inclusion $(Y, x_0) \to (X \lor Y, X)$ induces an isomorphism. This map has an obvious left inverse $(X \lor Y, X) \to (Y, x_0)$ that contracts $X$ to a point. By easy categorical considerations, this map also must induce an isomorphism in homology.
Now consider the long exact sequence for pair $(X \lor Y, X)$:
$ \cdots \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to \tilde{H}_{n-1}(X) \to \cdots$
The map $\tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$ is induced by inclusion $(X, x_0) \to (X \lor Y, x_0)$. There's an obvious left inverse to this inclusion, a map $(X \lor Y, x_0) \to (X, x_0)$ that contracts $Y$ in $X \lor Y$ to a point. As this map has a left inverse, the map induced in homology also has a left inverse, so it's a monomorphism, and we have an exact sequence $0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y)$
Similarly, consider $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$. As this map is also induced by inclusion $(X \lor Y, x_0) \to (X \lor Y, X)$, and from previous consideration we know that the contraction $(X \lor Y, X) \to (Y, x_0)$ induces an isomorphism, we consider a composition of these two. This map has a right inverse given by inclusion $(Y, x_0) \to (X \lor Y, x_0)$, which induces a left inverse for a map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(Y)$, which necessarily is an epimorphism. Now compose it with the inverse isomorphism $\tilde{H}_n(Y) \to \tilde{H}_n(X \lor Y, X)$ to obtain that $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is also epimorphism, and thus we get an exact sequence $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$. Remember that we composed it with the isomorphism inverse to the one $\tilde{H}_n(X \lor Y, X) \to \tilde{H}_n(Y)$ we used earlier, so here in this exact sequence, the map $\tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X)$ is the same as in long exact sequence.
Finally, we connect these two exact sequences to get a short exact sequence:
$0 \to \tilde{H}_n(X) \to \tilde{H}_n(X \lor Y) \to \tilde{H}_n(X \lor Y, X) \to 0$
As the first map is induced by inclusion and it has a right inverse we mentioned earlier, we obtain that this s.e.s. split, so $\tilde{H}_n(X \lor Y) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(X \lor Y, X) \simeq \tilde{H}_n(X) \oplus \tilde{H}_n(Y)$.
Note that additivity axiom was not used.