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Let $A_k=RP^2\sharp RP^2\sharp \cdots \sharp RP^2$ be a connected sum of $k$ copies of real projective space. With coefficients in $\mathbb{Z}$, it is clear $H_n(A_k)=0$ when $n\geq2$ and $H_0(A_k)=\mathbb{Z}$. However I'm trying to find $H_1(A_k)$.

After searching with google,I found that $H_1(A_k)=\mathbb{Z}^{k-1}\oplus \mathbb{Z}/2\mathbb{Z}$. and I think induction can deal with this. I tried to put $U=A_n$ with one point deleted and $V=RP^2$ with one point deleted in Mayer sequence $H_1(U\cap V)\to H_1(U)\oplus H_1(V)\to H_1(A_{n+1})$ but failed because I don't know what shape $U,U\cap V$ reformation retract to. what U and V should I put here?

Could you explain how to do this step-by-step? (like what theorems applied here so I can review it myself)

As far as I remember, my professor's already taught about homology group, Mayer-Vietoris, CW-complex, Euler characteristic...etc (but not cohomology)

Thanks for your help.

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    $\begingroup$ The standard tool for computing homology of connected sums is the Mayer-Vietrois sequence. See if you can compute $H_1(A_2)$ using it. $\endgroup$ – Braindead Jan 27 '14 at 2:39
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    $\begingroup$ As an alternative to the above use of Mayer-Vietoris, you can also find the fundamental group of $A_k$ using your favourite method (Van-Kampen decomposition/covering spaces) and then recall that $H_1(X)\cong\pi_1(X)^{ab}$. That is, the first homology group is equal to the abelianisation of the fundamental group of the space $X$ (for $X$ path-connected). $\endgroup$ – Dan Rust Jan 27 '14 at 15:13
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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.

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You can try drawing the polygon related to that connected sum, and use Mayer-Vietoris with $U$ equals your space minus a point, and $V$ a disk containing that point and strictly contained in your space. Then $U \cap V$ is $S^1$ and you should have no problems in completing the computation.

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