# Non-cohomological proof that a quasi-coherent sheaf over an affine scheme is quasi-flasque

Let $\mathcal F$ be a quasi-coherent sheaf over an affine scheme $X$. Let $0 \rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$ be an exact sequence of sheaves on $X$, where $\mathcal G$ and $\mathcal H$ are sheaves of abelian groups on $X$. Can we prove that $\Gamma(X, \mathcal G) \rightarrow \Gamma(X, \mathcal H)$ is surjective without using cohomology?

Remark Here's the definition of a quasi-flasque sheaf.

I'm kind of curious why you wouldn't want to use cohomology. Ok here's what I think (on very little sleep), if $\mathcal{G}$ and $\mathcal{H}$ are $\mathcal{O}_X$-modules, then the result is Hartshorne II.5.6 so give both of the latter sheaves the trivial module structure, i.e. $ag=g$ for $a\in \mathcal{O}_X(U)$ and $g\in \mathcal{G}(U)$ and then check that these give $\mathcal{O}_X$-modules and that the sequence is exact as $\mathcal{O}_X$-modules and if that's true, then you are done.

Edit: Ok, after getting some sleep, I looked over this again, and this definition from Wolfram must be mistaken http://imgur.com/iNUmgJJ by modul axioms (specifically, can $(a + a)g = g = ag + ag = g + g$?) so using II.5.6 wouldn't work unless you assume $G$ and $H$ are $O_X$-modules or somehow adapt that proof..

• Dear Aegbert, I don't think the fact you mentioned helps much. Please notice that $\mathcal G$ and $\mathcal H$ are not quasi-coherent $\mathcal O_X$-modules. Regards, Apr 2, 2014 at 7:28
• I misread, I thought they were all QC. Apr 2, 2014 at 7:34
• I mean it's still a proof of the fact, it's just using a little bit of cohomology... maybe you can show that it suffices to consider QC's Apr 2, 2014 at 7:46
• Ok, assuming they are all $O_X$-modules we can do it, this is Hartshorne II.5.6. Apr 2, 2014 at 8:22
• @aegbert There's no such thing as a "trivial module structure" – there are axioms that need to be satisfied! Apr 2, 2014 at 16:50

Notation Let $X$ be a topological space, $(U_i)_{i\in I}$ an open cover of $X$. Let $(i_0,\cdots, i_p) \in I^{p+1}$, where $p \ge 0$ is an integer. We denote $U_{i_0}\cap \cdots \cap U_{i_p}$ by $U_{i_0 \cdots i_p}$.

Proof of the Proposition Let $\psi\colon \mathcal F \rightarrow \mathcal G$ and $\phi\colon \mathcal G \rightarrow \mathcal H$ be the morphisms making the sequence $0 \rightarrow \mathcal F \rightarrow \mathcal G \rightarrow \mathcal H \rightarrow 0$ exact. Let $s\in \Gamma(X, \mathcal H)$. Since the morphism $\phi\colon \mathcal G \rightarrow \mathcal H$ is surjective and $X$ is quasi-compact, there exists a finite open cover $(U_i)_{i\in I}$ of $X$, where each $U_i = D(g_i)$ is a principal open subset, and a section $t_i \in \Gamma(U_i, \mathcal G)$ such that $\phi(t_i) = s|U_i$ for each $i \in I$. Since $\phi(t_i|U_{ij} - t_j|U_{ij}) = 0$, there exists $u_{ij} \in \Gamma(U_{ij}, \mathcal F)$ such that $\psi(u_{ij}) = t_i|U_{ij} - t_j|U_{ij}$ for each $(i,j) \in I^2$. By the following Lemma 1, there exists a section $u_i \in \Gamma(U_i, \mathcal F)$ for each $i \in I$ such that $u_{ij} = u_j|U_{ij} - u_i|U_{ij}$ for each $(i, j) \in I^2$. Then $t_i|U_{ij} - t_j|U_{ij} = \psi(u_j|U_{ij} - u_i|U_{ij})$. Hence $t_i|U_{ij} + \psi(u_i|U_{ij}) = t_j|U_{ij} + \psi(u_j|U_{ij})$. Therefore there exists $t \in \Gamma(X, \mathcal G)$ such that $t|U_i = t_i|U_i + \psi(u_i|U_i)$. Then $\phi(t) = s$. This completes the proof. QED

Lemma 1 Let $X$ be an affine scheme, $\mathcal F$ a quasi-coherent $\mathcal O_X$-module. Let $(U_i)_{i\in I}$ be a finite open cover of $X$, where each $U_i = D(g_i)$ is a principal open subset. Suppose a section $f_{ij} \in \Gamma(U_{ij}, \mathcal F)$ is given for each $(i, j) \in I^2$ such that $f_{ii} = 0, f_{ij} = -f_{ji}$. Furthermore suppose $f_{jk}|U_{ijk} - f_{ik}|U_{ijk} + f_{ij}|U_{ijk} = 0$ for each $(i, j, k) \in I^3$. Then there exists a section $f_i \in \Gamma(U_i, \mathcal F)$ for each $i \in I$ such that $f_{ij} = f_j|U_{ij} - f_i|U_{ij}$ for each $(i, j) \in I^2$.

Proof: Since $I$ is finite, there exists an integer $r \ge 1$ such that $$f_{ij} = \frac{x_{ij}}{(g_i g_j)^r}$$, $x_{ij} \in \Gamma(X, \mathcal F)$ for all $(i, j) \in I^2$.

Then $$\frac{g_i^r x_{jk}}{g_k^r (g_i g_j)^r}|U_{ijk} - \frac{g_j ^ rx_{ik}}{g_k^r (g_i g_j)^r}|U_{ijk} + \frac{g_k^r x_{ij}}{g_k^r(g_i g_j)^r}|U_{ijk} = 0$$ for each $(i, j, k) \in I^3$.

Hence there exists $l \ge 1$ such that

$$\frac{g_k^l g_i^r x_{jk}}{(g_i g_j)^r}|U_{ij} - \frac{g_k^l g_j ^ rx_{ik}}{(g_i g_j)^r}|U_{ij} + \frac{g_k^{r+l} x_{ij}}{(g_i g_j)^r}|U_{ij} = 0$$ for each $(i, j, k) \in I^3$.

Since the open subsets $D(g_k^{r+l})$ cover $X$, $1 = \sum_k h_k g_k^{r+l}$, $h_k \in \Gamma(X, \mathcal O_X)$.

Let $$f_i = \sum_k h_k g_k^l \frac{x_{ki}}{g_i^r} \in \Gamma(U_i, \mathcal F)$$ for each $i \in I$.

Then $$f_i|U_{ij} = \sum_k h_k g_k^l g_j^r \frac{x_{ki}}{(g_i g_j)^r}|U_{ij}$$.

$$f_j|U_{ij} = \sum_k h_k g_k^l g_i^r \frac{x_{kj}}{(g_j g_j)^r}|U_{ij}$$.

Hence $$f_j|U_{ij} - f_i|U_{ij} = \sum_k h_k g_k^l (g_i^r \frac{x_{kj}}{(g_i g_j)^r}|U_{ij} - g_j^r \frac{x_{ki}}{(g_i g_j)^r}|U_{ij}) = \sum_k h_k g_k^{r+l}\frac{x_{ij}}{(g_i g_j)^r} = f_{ij}$$. QED

• This appears to be the proof that one obtains by unfolding the definition of the first Čech cohomology group and showing that it vanishes for quasi-coherent sheaves. Apr 4, 2014 at 15:26
• @ZhenLin Maybe, but at least you don't need cohomological machinery like a cohomology sequence or flasque or injective resolutions here. Apr 4, 2014 at 15:54
• This proof is used in my answer to this question. Apr 6, 2014 at 1:18