# Apparent contradiction using Mayer-Vietoris for Sheaf cohomology

Trying to solve Exercise 2.7 b) of Chapter III of Hartshorne's Algebraic Geometry I got stucked about an apparent contradiction.

The exercise asks to prove that $H^1(S^1, \mathcal{R})=0$, where $S^1$ is the circle with its usual topology and $\mathcal{R}$ is the sheaf of continuous real-valued functions.

I thought to apply the Mayer-Vietoris sequence for the cohomology with support to the closed subsets of $S^1$:

$U:=\{(cos\theta, sen\theta): \theta \in [o,\pi]\},$

$V:=\{(cos\theta, sen\theta): \theta \in [\pi,2\pi]\},$

Their intersection is the pair of points $\{(1,0),(-1,0)\}$.

We observe that $\mathcal{R}$ is a flasque sheaf (i.e. if $U\subseteq V$ are open sets then the restriction map $\mathcal{R}(V) \rightarrow \mathcal{R}(U)$ is a surjection, thanks to Exercise 2.3 c) of the same Chapter, all the cohomology group of grade greater than $0$ computed on a flasque sheaf are trivial. This would solve the exercise by itself but just keep thinking in this direction for a moment...). And we recall that $H^0_Y(X,\mathcal{F})=\Gamma_Y(X,\mathcal{F})$ we get the (short) exact sequence:

$0 \rightarrow \Gamma_{U \cap V}(S^1, \mathcal{R}) \rightarrow \Gamma_U(S^1, \mathcal{R})\oplus\Gamma_V(S^1, \mathcal{R}) \rightarrow \Gamma(S^1, \mathcal{R}) \rightarrow 0.$

Taking a closer look to the groups involved I noticed that $\Gamma_{U \cap V}(S^1, \mathcal{R})\simeq 0$ because the germ of a continuous function on a point which is zero outside of that point must be zero also on the point, by continuity. For the other terms $\Gamma_U(S^1, \mathcal{R}) \simeq \Gamma_V(S^1, \mathcal{R}) \simeq \mathcal{C}([0,1])_{\partial=0}$ the group of continuous functions on $[0,1]$ which are zero on the boundary. And $\Gamma(S^1, \mathcal{R}) \simeq \mathcal{C}([0,1])_{per}$ the group of continuos functions on $[0,1]$ such that $f(0)=f(1)$.

By the sequence above we deduce that $\mathcal{C}([0,1])_{per} \simeq \mathcal{C}([0,1])_{\partial=0}\oplus \mathcal{C}([0,1])_{\partial=0}$. But really I don't see it (I tried to do something with Borsuk-Ulam theorem but the matter is I don't think it is true).

Can someone please explain where my error lies? Or, otherwise, how the groups as above are isomorphic?

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It is not true that $\mathcal R$ is flasque: the function $1/(x-1)$ on the circle minus $(0,0)$ cannot be extended continuously to the circle .
However $H^1(S^1,\mathcal R)=0$ because the sheaf $\mathcal R$ is soft.
Note however that for the constant sheaf $\underline{\mathbb R}$ we have $H^1(S^1,\underline {\mathbb R})=\mathbb R$ because on a paracompact space cohomology of a constant sheaf coincides with singular cohomology and $H^1_{\operatorname {sing}}(S^1,\underline {\mathbb R})=\mathbb R$

Remark
Note that in order to prevent any misunderstanding with the constant sheaf $\underline {\mathbb R}$ it might be safer to use the much more usual notation $\mathcal C_X$ (rather than $\mathcal R$) for the sheaf of real valued continuous functions on a topological space $X$.

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Thank you for your help! Actually my edition of Hartshorne's (ISBN 0-387-90244-9) clearly asks to prove $H^1(S^1,\mathcal{R})=0$, it is probably a typo corrected in other editions. – Giovanni De Gaetano Feb 22 '12 at 11:40
Dear @Student: ah, I see. Sorry for having said you had misread the exercise. I have edited my post to remove that unfair assertion. – Georges Elencwajg Feb 22 '12 at 13:18
Dear Georges, if we compute Cech cohomology, we find $H^1(S^1,\mathcal R)=0$: math.stackexchange.com/a/1535053/284566. How do we reconcile this with your claim? – user4571 Nov 18 '15 at 15:02
Dear @Patrick: you are absolutely right! Hartshorne's not so great notation $\mathcal R$ instead of $\mathcal C$ for the sheaf of continuous functions confused me and I answered as if $\mathcal R$ meant the constant sheaf $\underline {\mathbb R}$ with fiber $\mathbb R$. I have modified my answer which should now be unambiguous . Thanks a lot for your attention. – Georges Elencwajg Nov 18 '15 at 20:17