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.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.