Proving that $\tilde{H}^i(\mathbb R P^2 \vee S^3; \mathbb Z_2) = \tilde{H}^i(\mathbb R P^2; \mathbb Z_2) \oplus \tilde{H}^i(S^3; \mathbb Z_2)$

Question

I want to prove the following statement:

$\tilde{H}^i(\mathbb R P^2 \vee S^3; \mathbb Z_2) = \tilde{H}^i(\mathbb R P^2; \mathbb Z_2) \oplus \tilde{H}^i(S^3; \mathbb Z_2)$

but I am not sure how to prove it, I am guessing that it should be proved using Mayer Vietoris sequence as I think I have seen something similar to this but for homology before( I am not sure where did I saw it though).

Could anyone show me the details of the proof please?

Here is my trial:

Here are my A, B and X. $A$ is the upper hemisphere of $\mathbb R P^2$ and $B$ is the lower hemisphere of $S^3$ and $X$ is the figure on the right below (their union with an intersection strip between A and B):

[![enter image description here][1]][1]

Now I know that $\tilde{H^i}(\mathbb RP^{n}, \mathbb Z/2\mathbb Z) = \mathbb Z/2\mathbb Z$ for $0 < i \leq n$ and 0 when $i=0.$ Then I can write the Mayier Vietoris sequence as follows:

$... \tilde{H^2}(A \cap B) \xrightarrow{\phi} \tilde{H^2}(A) \oplus \tilde{H^2}(B) \xrightarrow{\psi} \tilde{H^2}(X) \xrightarrow{\partial} \tilde{H^1}(A \cap B) \xrightarrow{\phi} \tilde{H^1}(A) \oplus \tilde{H^1}(B) \xrightarrow{\psi} \tilde{H^1}(X) \xrightarrow{\partial}0$

but then why always we must have the reduced homology of the intersection is 0 and why the space $X$ can be considered as $\mathbb R P^2 \vee S^3$?

Could someone explain this to me please? [1]: https://i.stack.imgur.com/AkZAG.png

Answer 1

You need to be able to show that, for a point in each space involved in the wedge, there is a neighborhood of that point which deformation retracts to that point. This is true for any simplicial complex (any $0$-simplex has such a neighborhood) or more generally any CW complex. If $(X,v)$ and $(Y,w)$ are two such pairs of (space, nice point), then form the wedge by gluing $v$ to $w$. Let $V \subseteq X$ and $W \subseteq Y$ be neighborhoods which retract to $v$ and $w$, respectively. Now apply Mayer-Vietoris to $A = X \cup W$ and $B = V \cup Y$. (Think of $X \cup W$ as being $X$ with a little fluff added: $X \cup W$ will be homotopy equivalent to $X$ and hence have the same homology as $X$, but constructed so that $X$ is in the interior of $X \cup W$, as required by the statement of Mayer-Vietoris.) Their intersection is $V \vee W$, which deformation retracts to the wedge point $v=w$, and so is contractible. Therefore Mayer-Vietoris will give $$ \tilde{H}_n(X \vee Y) \cong \tilde{H}_n(X) \oplus \tilde{H}_n(Y) $$ for each $n$.

Alternatively: first prove that if $X \amalg Y$ is the disjoint union, then $H_n(X \amalg Y) \cong H_n(X) \oplus H_n(Y)$. In fact, prove a relative version of this: $$ H_n(X \amalg Y, \{v\}\amalg \{w\}) \cong H_n(X, v) \oplus H_n(Y, w). $$ With the assumption above about nice neighborhoods of $v$ and $w$, we get isomorphisms $$ H_n(X \amalg Y, \{v\}\amalg \{w\}) \cong \tilde{H}_n(X \amalg Y/ \{v\}\amalg \{w\})) = \tilde{H}_n(X \vee Y) \\ H_n(X, v) \cong \tilde{H}_n(X) \\ H_n(Y, w) \cong \tilde{H}_n(Y) $$ The first of these happens because when we take the quotient, we are collapsing $\{v\}\amalg \{w\}$ to a single point — we are constructing the wedge.