Let $X$ be the space consisting of two tori glued along the first circle: I interpret the space as $X=S^1_{(1)}\times S^1_{(2)}\sqcup S^1_{(3)}\times S^1_{(4)} / (S^1_{(1)}\sim S^1_{(3)})$. I want to compute its cohomology (over, say, the integers). I have two ideas for this:
- Use the long exact sequence of the quotient on $(X,S^1_{(1)})$
For the first approach, I have two preliminary steps that I'm not sure of: first, the quotient of the torus $S^1\times S^1/S^1\times \{e\}$ by its first circle is just the circle. If that's the case, then the quotient $X/S^1$ is the wedge sum of two circles $S^1\vee S^1$. But then the LES of the quotient is:
$$0\to \tilde H^1(S^1\vee S^1)\to \tilde H^1(X)\to \tilde H^1(S^1)\to \tilde H^2(S^1\vee S^1)\to \tilde H^2(X)\to \tilde H^2(S^1)\to \tilde H^3(S^1\vee S^1)\to \tilde H^3(X)\to \tilde H^3(S^1)\to 0$$
But $\tilde H^n(S^1\vee S^1)= \tilde H^n(S^1)\oplus \tilde H^n(S^1)$, hence the LES becomes:
$$0\to \mathbb Z^2\to \tilde H^1(X)\to\mathbb Z\to 0 \to \tilde H^2(X)\to 0\to 0\to \tilde H^3(X)\to 0\to 0$$
Which looks completely wrong, since we get $\tilde H^1(X)=\mathbb Z^3$ and $0$ otherwise.
- Use Mayer-Vietoris on $X$ by taking as my two subspaces each torus.
For the second approach, the intersection of the tori in $X$ is just $S^1$, so Mayer Vietoris gives:
$$0\to H^0(X)\to H^0(A)\oplus H^0(B)\to H^0(A\cap B)\to H^1(X)\to H^1(A)\oplus H^1(B)\to H^1(A\cap B)\to H^2(X)\to H^2(A)\oplus H^0(B)\to H^2(A\cap B)\to 0$$ Which gives
$$0\to \mathbb Z\to \mathbb Z^2\to \mathbb Z\to H^1(X)\to \mathbb Z^4\to \mathbb Z\to H^2(X)\to \mathbb Z^2\to 0\to 0$$
After some playing around this last sequence then would (could?) imply that $H^1\cong \mathbb Z^4$ and $H^2\cong \mathbb Z$, which looks slightly more reasonable.
What went wrong?