# Computing de Rham cohomology group $H^1(\mathbb{RP}^2\#\mathbb{RP}^2)$

I am trying to compute the de Rham cohomology group $$H^p(\mathbb{RP}^{n+1}\#\mathbb{RP}^{n+1})$$ and I am stuck at computing $$H^1(\mathbb{RP}^2\#\mathbb{RP}^2)$$. ($$\#$$ stand for the connected sum)

Let $$U\simeq \mathbb{RP^2\setminus \{p\}}$$ and $$V\simeq \mathbb{RP}^2\setminus \{q\}$$ such that $$M=\mathbb{RP}^2\#\mathbb{RP}^2=U\cup V$$ and $$U\cap V\simeq S^1$$. Then by using Mayer Vietoris, I get \begin{align*} 0&\xrightarrow{a} H^0(M)=\mathbb{R}\xrightarrow{b} H^0(U)\oplus H^0(V)=H^0(\mathbb{RP}^1)\oplus H^0(\mathbb{RP}^1)=\mathbb{R}\oplus \mathbb{R}\xrightarrow{c} H^0(S^1)=\mathbb{R} \\& \xrightarrow{d} H^1(M)\xrightarrow{e} H^1(U)\oplus H^1(V)=H^1(\mathbb{RP}^1)\oplus H^1(\mathbb{RP}^1)=\mathbb{R}\oplus \mathbb{R} \xrightarrow{f} H^1(S^1)=\mathbb{R}\\&\xrightarrow{g} H^2(M)\xrightarrow{h} H^2(U)\oplus H^2(V)=H^2(\mathbb{RP}^1)\oplus H^2(\mathbb{RP}^1)=0. \end{align*}

But the fact that it is exact sequence does not give me specific feature of $$H^1(M)$$ and $$H^2(M)$$. I have computed two possibilities.

Note that $$d$$ is zero map, so $$e$$ is injective. So only possible choice of $$H^1(M)$$ is $$0$$, $$\mathbb{R}$$ and $$\mathbb{R}\oplus \mathbb{R}$$. But $$H^1(M)$$ cannot be $$0$$ since otherwise $$f$$ is injective and it is impossible.

Then observe the following two cases.

(1) If $$H^1(M)$$ is $$\mathbb{R}$$, $$g$$ is surjective zero map which means $$H^2(M)=0$$.

(2) If $$H^1(M)$$ is $$\mathbb{R}\oplus \mathbb{R}$$, $$f$$ is zero map so $$g$$ is isomorphism. Thus, $$H^2(M)=\mathbb{R}$$.

As we can see, either ways does not make any contradiction.

I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)

• The top cohomology of a non-orientable manifold must be $0$. – Cheerful Parsnip Dec 24 '18 at 23:41
• @CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from? – Lev Ban Dec 24 '18 at 23:42
• See the computation here: math.stackexchange.com/questions/187413/… – anomaly Dec 24 '18 at 23:43
• @LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum. – Cheerful Parsnip Dec 25 '18 at 0:04
• Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle. – Cheerful Parsnip Dec 25 '18 at 0:08