# Sheaf of a Closed Subset

I’ve been given the following definition:

Let $$(X,\mathcal{O}_X)$$ be a ringed space which is locally isomorphic to an affine algebraic variety, and $$Y\subseteq X$$ be closed. Then for an open $$V\subseteq Y$$, set \begin{align*} \mathcal{O}_{0,Y}(V)=\{f:V\to k\mid{}&\exists U\subseteq X\text{ open such that }U\cap Y=V\\ &\text{and } g\in\mathcal{O}_X(U)\text{ such that } g\vert_V=f\} \end{align*} This defines a presheaf $$\mathcal{O}_{0,Y}$$ on $$Y$$, but not, in general, a sheaf.

However I’m struggling to come up with an example where this fails to be a sheaf.

I thought I'd found a counterexample with $$X=\mathbb{C}^2$$, $$Y=V(xy)$$, $$U=D(x)\cap Y$$ and $$V=D(y)\cap Y$$. Then $$U\cap V=\varnothing$$, and so if $$\mathcal{O}_{0,Y}$$ were a sheaf, then we would be able to glue to make a function on $$U\cup V$$ which is say $$1$$ on $$U$$ and $$-1$$ on $$V$$.

I can show that we can't get such a function from gluing two functions on $$D(x)$$ and $$D(y)$$, but we can take $$\frac{x+y}{x-y}$$ on $$D(x-y)$$ to give the required function. Then it isn't enough to just check the 'obvious' open cover, and I haven't yet been able to find a counterexample which works for every one.

Any help would be much appreciated.

• I haven't checked fully, but perhaps $X$ is the affine line with doubled origins and $Y$ is the closed subsets consisting of the two origins $\{o_1,o_2\}$. My gut says that $\mathcal{O}(o_1)\cong\mathcal{O}(o_2)\cong \mathcal{O}(Y)\cong k$ but if it were a sheaf you'd have that $\mathcal{O}(Y)\cong k^2$. Just a thought. Apr 23, 2019 at 2:39

This is kind of messy and could probably be written better, but I think this works:

Take $$X$$ to be the line with "double origin", that is, we define $$X$$ by gluing two copies of $$\Bbb A^1$$ together along the open subset $$\Bbb A^1\smallsetminus\{0\}$$ (with identity as the isomorphism we identify these open subsets). Because I will want to refer to the copies of $$\Bbb A^1$$, let $$X_0$$ and $$X_1$$ denote our two copies of $$\Bbb A^1$$, which are now naturally identified with open subsets of $$X$$, and let $$O_0,O_1$$ denote the two "origins", so $$O_i\in X_i$$.

Note now that $$\{O_0\}=X\smallsetminus X_1$$, so $$O_0$$ is a closed point, and similarly $$O_1$$ is a closed point. Take $$Y:=\{O_0,O_1\}$$, which is then a closed subset and the subspace topology is the discrete topology, so $$\{O_0\}$$ and $$\{O_1\}$$ are open subsets of $$Y$$. Then we can look at constant map $$f_i:\{O_i\}\to k$$ sending $$O_i\mapsto i$$, and it's not hard to check this gives us an element of $$\mathcal O_{0,Y}(\{O_i\})$$.

We claim $$f_0,f_1$$ will not glue to an element of $$\mathcal O_{0,Y}(Y)$$. If they do, say $$f\in\mathcal O_{0,Y}(Y)$$, then by definition there is an open subset $$U$$ of $$X$$ which contains $$Y$$ and an element $$g\in\mathcal O_X(U)$$ such that $$g|_Y=f$$. But by definition of a sheaf, because $$X_0$$ and $$X_1$$ cover $$X$$, an element $$g\in\mathcal O_X(U)$$ is the same thing as a pair of elements $$g_i\in\mathcal O_X(U\cap X_i)$$ for $$i=0,1$$ which are equal on the intersection.

Now, because $$X_0$$ is really just $$\Bbb A^1$$, the complement of $$U\cap X_0$$ in $$X_0$$ is a finite set of points $$a_1,\dots,a_m$$. Therefore $$g_0$$ is just a ratio of two polynomials, the denominator of which does not vanish at any $$a_i$$, and we can write $$g_1$$ in the same way. But these two agree on the overlap, which consists of infinitely many points (we should assume we are over $$\Bbb C$$ or any other algebraically closed field here), so you can try to use this to conclude that the expressions which define $$g_0$$ and $$g_1$$ are rational functions are in fact equal (you should use the fact that if two polynomials agree on an infinite subset, then they are equal), so $$g_0$$ and $$g_1$$ should be equal everywhere they are defined (and in particular, on $$Y$$). But we also have

$$g_i|_{\{O_i\}}=(g|_{X_i})|_{\{O_i\}}=g|_{\{O_i\}}=(g|_Y)|_{\{O_i\}}=f|_{\{O_i\}}=f_i,$$

and because $$f_0$$ and $$f_1$$ take different values at our two origins this is impossible.

• Lol, I think we posted our comment/answer literally simultaneously. Apr 23, 2019 at 2:41