Homology of connected sum of real projective spaces

Let $X$ be the first $(k - 1)$ factors in $A_k$ and $Y$ be the last factor. Let $S^1$ be the circle that connects $X$ with $Y$. Let $W$ be a neighborhood of $S^1$ that deformation retracts onto $S^1$. Define $U = X \cup W$, $V = Y \cup W$.

Using he Mayer-Vietoris sequence for reduced homology, we have the following exact sequence: $$ \tilde H_1(U \cap V) \xrightarrow{\varphi} \tilde H_1(U) \oplus \tilde H_1(V) \xrightarrow{\psi} \tilde H_1(A_k) \rightarrow 0 $$

By considering the fundamental polygon for $\Bbb R \textrm P^2$, one can see that $U$ deformation retracts onto the wedge sum of $(k - 1)$ circles. Similarly, $V$ deformation retracts onto a circle. Hence $\tilde H_1(U) \cong \Bbb Z^{k-1}$, $\tilde H_1(V) \cong \Bbb Z$. Since $U \cap V = W$ deformation retracts onto $S^1$, we have $\tilde H_1(U \cap V) \cong \Bbb Z$.

What's left is to find $\ker \psi = \operatorname{im} \varphi$ and apply the first isomorphism theorem.

From the definition of the Mayer-Vietoris sequence, $\varphi = (i_*, j_*)$ where $i : U \cap V \hookrightarrow U$, $j : U \cap V \hookrightarrow V$ are the inclusion maps.

Since the generator of $\tilde H_1(U \cap V)$ goes twice around each generator of the circles of $U$, we have $$ i_*(1) = \underbrace{(2, \ldots, 2)}_{(k - 1) \text{ times}}. $$

Similarly, $j_*(1) = 2$.

Hence $$ \ker \psi \cong \underbrace{(2, \ldots, 2)}_{k \text{ times}} \Bbb Z.$$

By an application of the first isomorphism theorem, we have $$ \tilde H_1(A_k) \cong \left(\tilde H_1(U) \oplus \tilde H_1(V)\right) / \ker \psi = \Bbb Z^{k-1} \oplus \Bbb Z_2. $$

Note that one can prove the following more general result:

If $M_1$ and $M_2$ are closed manifolds then there are isomorphisms $H_i(M_1 \# M_2) \cong H_i(M_1) \oplus H_i(M_2)$ for $0 < i < n$, with one exception: If both $M_1$ and $M_2$ are nonorientable, then $H_{n−1}(M_1 \# M_2)$ is obtained from $H_{n−1}(M_1) \oplus H_{n−1}(M_2)$ by replacing one of the two $\Bbb Z_2$ summands by a $\Bbb Z$ summand.

The proof is similar to what I have above, but requires some manifold theory.