$H^0(X,\mathcal{F})=\mathcal{F}(X)$

Question

Let $X$ a topological space and $\mathcal{F}$ be a sheaf. I have to prove that $H^0(X,\mathcal{F})=\mathcal{F}(X)$.

I have the cochain $0\rightarrow \mathcal{F}_0(X)\rightarrow \mathcal{F}_1(X)\rightarrow...$, with $\mathcal{F}_0=\bar{\mathcal{F}}$, $\mathcal{F}_1=\bar{\mathcal{F}_1^{'}}$ where $\mathcal{F_1^{'}}=\bar{\mathcal{F}}/\mathcal{F} $.

$\bar{\mathcal{F}}(U) =\prod_{x\in U} \mathcal{F}_x$

$H^0(X,\mathcal{F}):= Ker(\mathcal{F}_0(X)\rightarrow \mathcal{F_1}(X))$

Now I consider the projection $\pi:\bar{\mathcal{F}}(X)\rightarrow\bar{\mathcal{F}}(X)/\mathcal{F}(X)$ and the inclusion $i:\bar{\mathcal{F}}(X)/\mathcal{F}(X)\rightarrow \bar{\mathcal{F_1^{'}}}(X)$.

Therefore I obtain

$i\circ\pi$: $\mathcal{F}_0(X)\rightarrow \mathcal {F_1}(X) $.

$H^0(X,\mathcal{F})= Ker (i\circ \pi)= \mathcal{F}(X)$ because $i$ is injective.

It's right?

Answer 1

Remember that to compute $H^k(X,F)$ you have to choose an injective resolution of F $$0\to F\overset{i}\to F_1 \overset{f_1}\to F_2 \to ...$$ and then consider the exact sequence $$F_1(X) \overset{f_1(X)}\to F_2(X) \overset{f_2(X)}\to F_3(X) \to ...$$ Then, we have $$H^k(X,F) =\text{ker}(f_{k+1}(X))/\text{im}(f_{k}(X)).$$ So in your case, we have simply $$H^0(X,F) = \text{ker}(f_1(X)) = \text{im}(i(X))=F(X).$$

Answer 2

What cohomology are you considering? If it is the $\check{\text{C}}$ech cohomology, the situation (for cohomology groups in general degrees) is more delicate.

First of all, you need a covering of your topological space $X$. Let $\mathfrak{U} :=\{U_i\}_{i\in I}$ denote such a covering. Then consider the chain complex $$ 0 \to \prod_{i\in I}\mathcal{F}(U_i) \stackrel{\partial}\to \prod_{i,j \in I}\mathcal{F}(U_i\cap U_j) \stackrel{\partial}{\to}\dots. $$ In general the $k$-th group in this chain complex is defined by $$ C^k(\mathfrak{U};\mathcal{F}):= \prod_{i_1,\dots,i_{k+1}\in I} \mathcal{F}(U_{i_k} \cap \dots \cap U_{i_{k+1}}), $$ and the coboundary operator is given by $$ \partial: C^k(\mathfrak{U};\mathcal{F}) \to C^{k+1}(\mathfrak{U};\mathcal{F}), $$ $$ (\partial(f))_{i_1\dots i_{k+2}} = \sum_{j=1}^{k+2} (-1)^{j+1}\left. f_{i_1\dots \hat{i_j} \dots {i_{k+2}}}\right|_{U_{i_1} \cap\dots\cap \hat{U_{i_j}}\cap\dots \cap U_{i_k}}. $$ To clarify the notation a little bit, elements of $C^k(\mathfrak{U};\mathcal{F})$ are sequences of elements indexed by $I^{k+1}$, so the formula above gives a certain component of such a sequence, whereas $f$ denotes the whole sequence. The hat indicates that the index is omitted. The sheaf cohomology (corresponding to the cover $\mathfrak{U}$) is defined by $$ H^k(\mathfrak{U};\mathcal{F}):=\frac{\ker(\partial:C^k(\mathfrak{U};\mathcal{F})\to C^{k+1}(\mathfrak{U};\mathcal{F}))}{\text{im}(\partial: C^{k-1}(\mathfrak{U};\mathcal{F}) \to C^k(\mathfrak{U};\mathcal{F})}. $$ Note that this definition certainly does depend on the cover, and in general, one must take a direct limit which is for the moment very difficult to compute. But fortunately, this is not the case for $k =0$.

Obviously the subspace that should be quotient out in the definition is zero, so we only need to find the kernel of $$ \partial: C^0(\mathfrak{U};\mathcal{F}) \to C^1(\mathfrak{U};\mathcal{F}). $$ But if $f = (f_i)_{i\in I}\in C^0(\mathfrak{U};\mathcal{F})$, such that $$ \partial(f) = 0, $$ then $$ (\partial(f))_{ij} = 0 $$ for all $i,j \in I$. Using the definition, we find $$ \left.f_i\right|_{U_i\cap U_j} = \left.f_j\right|_{U_i\cap U_j}. $$ But by definition of a sheaf, we know there exists an $g\in \mathcal{F}(X)$, such that $$ \left.g\right|_{U_i} = f_i. $$ So elements of $$ \ker(\partial:C^0(\mathfrak{U};\mathcal{F}) \to C^1(\mathfrak{U};\mathcal{F}), $$ precisely correspond to global sections $g\in \mathcal{F}(X)$. This is true independent of the chosen cover $\mathfrak{U}$, the cohomology group $$ H^0(X;\mathcal{F}) = \ker(\partial:C^0(\mathfrak{U};\mathcal{F}) \to C^1(\mathfrak{U};\mathcal{F})) \cong \mathcal{F}(X). $$