Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am using the Mayer Vietoris sequence to compute the de Rham cohomology of a twice punctured plane.

I computed it first by choosing my open cover to be $U=R^2-{pt}$ and $V=R^2-{pt}$ (the points are different) with $U\cup V = R^2$ and $U \cap V$ the double punctured plane. Everything worked as expected.

Then, I tried computing it another way. Assume the missing points are (0,1) and (0,-1). I take $U = [(x,y): y > -1/2] $ with (0,1) missing. Similarly, I take $V = [(x,y): y < 1/2] $ with (0,-1) missing. Then, $ U \cup V $ is the double punctured plane, and $ U \cap V $ is the convex strip $ [-1/2 < y < 1/2] $ . But $H^{0}(U) \oplus H^{0}(V) = H^{1}(U) \oplus H^{1}(V) = R^2$ while $H^{0}(U \cap V) = R$ and $H^{1}(U \cap V) = 0$, so that the sequence is not exact: $ H^{0}(U \cap V) = R$ $\rightarrow$ $H^{1}(U \cup V)$ $\rightarrow$ $H^{1}(U) \oplus H^{1}(V) = R^2$ $\rightarrow$ $H^1(U \cap V) = 0$
What blunder am I making here?


share|cite|improve this question
Where in the sequence is the failure of exactness? – Nick D. Jan 23 '14 at 20:08
You're right. There's no problem – Guest Jan 23 '14 at 20:22
up vote 3 down vote accepted

There's no blunder. The MV sequence says this (where $M = U \cup V$), and I'm gonna be sloppy about formatting: $$ 0 \to H^0M \to H^0 U + H^0 V \to H^0(U \cap V) \to \\ H^1 (M) \to H^1 U + H^1 V \to H^1(U \cap V) \to \ldots $$ Let's fill in those groups: $$ 0 \to R \to R + R\to R\to \\ R+R \to R+R \to 0 \to \ldots $$ The maps themselves look like this: $$ 0 \to R : 0 \mapsto 0\\ R \to R + R :t \mapsto (t, t)\\ R + R\to R : (u,v) \mapsto u - v \\ R \to R + R :t \mapsto 0\\ R+R \to R+R : (u, v) \mapsto (u, v) \\ R+R \to 0 : (u, v) \mapsto 0. $$ That's exact at every step.

You might want to look at as well.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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