Sheaf Operations Exactness Ringed Spaces and Module Pullback - Examples

Because $j$ is the inclusion of $U$ into $X$, one has $j^{-1}(V)=V\cap U$.

Substituting step 1.1 into [F1] gives $(j_{\ast }\mathcal{F})(V)=\mathcal{F}(V\cap U)\mathrm{.}$ This is exactly the announced formula, and it is compatible with the restriction interpretation in [F2].

By [F1], the global section set of $j_{\ast }\underset{‾}{\mathbb{Z}}$ is $(j_{\ast }\underset{‾}{\mathbb{Z}})(X)=\underset{‾}{\mathbb{Z}}(U)\cong \mathbb{Z}\times \mathbb{Z},$ because $U$ has two connected components. Let $s$ be the section that is $1$ on both components.

The germ of $s$ is nonzero at every point of $U$, so its support is all of $U$. But $U=(-1,0)\cup (0,1)$ is not closed in $X=(-1,1)$, since its closure contains $0$. Therefore [F2] shows that $s\mathrm{\notin }(j_{!}\underset{‾}{\mathbb{Z}})(X)$.

Thus $s$ lies in $j_{\ast }\underset{‾}{\mathbb{Z}}$ but not in $j_{!}\underset{‾}{\mathbb{Z}}$, so the two sheaves are not equal. This refutes the statement.

Fix a point $y\in X$. If every neighbourhood of $y$ contains $x$, then [F1] shows that each stalk in the displayed skyscraper sequence is the original group at $y$, so the stalk sequence is exactly $0\to A^{\mathrm{\prime }}\to A\to A^{\mathrm{\prime }\mathrm{\prime }}\to 0.$ If some neighbourhood of $y$ omits $x$, then every smaller neighbourhood also omits $x$, so [F1] makes all three stalks equal to $0$ and the stalk sequence is $0\to 0\to 0\to 0.$ In either case the stalk sequence is exact.

The stalk sequence is therefore exact at every point, so [L1] implies that the skyscraper sequence is exact as a sequence of sheaves. By [L2], this is exactly the claimed short exact sequence of skyscraper sheaves.

Let $\mathcal{C}$ be the sheaf of continuous real-valued functions on $S^1$, and let $\underset{‾}{\mathbb{Z}}$ be the subsheaf of locally constant integer-valued functions. Let $\mathcal{Q}$ be the sheafification of the presheaf quotient $V\mapsto \mathcal{C}(V)\mathrm{/}\underset{‾}{\mathbb{Z}}(V)$. Every germ of $\mathcal{Q}$ is represented locally by a continuous function, so the canonical map $\mathcal{C}\to \mathcal{Q}$ is locally surjective and hence an epimorphism of sheaves.

Cover the circle by the two arcs $U_0=S^1\setminus \{(-1,0)\}$ and $U_1=S^1\setminus \{(1,0)\}$. Choose continuous angle functions ${\theta }_0:U_0\to (-1\mathrm{/}2,1\mathrm{/}2)$ and ${\theta }_1:U_1\to (0,1)$ with $e^{2\pi i{\theta }_i(z)}=z$. Their quotient classes agree on $U_0\cap U_1$, because the difference ${\theta }_1-{\theta }_0$ is locally constant integer-valued there. Thus they glue to a section $q\in \mathrm{\Gamma }(S^1,\mathcal{Q})$. If $q$ came from a global continuous real function $g$, then $g-{\theta }_i$ would be an integer-valued continuous function on the connected set $U_i$, hence constant. On the two connected components of $U_0\cap U_1$ this would force the same constant difference to be both $0$ and $1$, which is impossible. So $\mathrm{\Gamma }(S^1,\mathcal{C})\to \mathrm{\Gamma }(S^1,\mathcal{Q})$ is not surjective.

Therefore an epimorphism of sheaves can fail to become surjective on global sections. This is exactly the phenomenon asserted in [F1], so the statement is refuted.

The usual restriction of continuous functions makes $U\mapsto {\mathcal{C}}_X^0(U)$ a sheaf of commutative rings on $X$, so [F1] gives a ringed space $(X,{\mathcal{C}}_X^0)$.

Fix $x\in X$. By [F2], the stalk ${\mathcal{C}}_{X,x}^0$ consists of germs of continuous real-valued functions near $x$. The germs vanishing at $x$ form an ideal ${\mathfrak{m}}_x$. If a germ is not in ${\mathfrak{m}}_x$, it has a representative $g$ with $g(x)\mathrm{\ne }0$, so continuity makes $g$ nonzero on a smaller neighbourhood of $x$; hence $1\mathrm{/}g$ is continuous there and defines an inverse germ. Therefore the nonunits are exactly the germs in ${\mathfrak{m}}_x$, so ${\mathfrak{m}}_x$ is the unique maximal ideal. Thus ${\mathcal{C}}_{X,x}^0$ is a local ring, and [L1] shows that $(X,{\mathcal{C}}_X^0)$ is locally ringed.

Regard $(\{\ast \},A)$ and $(\{\ast \},K)$ as one-point ringed spaces. The assignment $y\mapsto x$ defines a ring homomorphism $A\to K$, so by [F1] it defines a morphism of ringed spaces $(\{\ast \},K)⟶(\{\ast \},A)\mathrm{.}$

The unique maximal ideal of $A$ is $(y)$ by [L1], while the unique maximal ideal of the field $K$ is $0$. Under the stalk map $A\to K$, the element $y$ goes to the nonzero element $x$, hence to a unit of $K$. Therefore the image of $(y)$ is not contained in $0$, so the stalk map is not local. By [F2], the morphism is not a morphism of locally ringed spaces.

This gives a morphism of ringed spaces between locally ringed spaces that is not locally ringed. The statement is false.

The sheaf ${\mathcal{O}}_Y^n$ is the finite direct sum of $n$ copies of ${\mathcal{O}}_Y$, with the case $n=0$ giving the zero sheaf. Since [L1] makes $f^{\ast }$ a left adjoint, it preserves these finite direct sums. Thus $f^{\ast }({\mathcal{O}}_Y^n)\cong (f^{\ast }{\mathcal{O}}_Y{)}^n\mathrm{.}$

Applying [F1] to $\mathcal{G}={\mathcal{O}}_Y$ gives $f^{\ast }{\mathcal{O}}_Y={\mathcal{O}}_X{\otimes }_{f^{-1}{\mathcal{O}}_Y}{f}^{-1}{\mathcal{O}}_Y\cong {\mathcal{O}}_X,$ because tensoring a module over a ring with the ring itself leaves the module unchanged. Substituting this into step 1.1 yields $f^{\ast }({\mathcal{O}}_Y^n)\cong {\mathcal{O}}_X^n\mathrm{.}$

On $U_i\cap U_j$, define ${\phi }_{ij}:{\mathcal{O}}_X{\mathrm{\mid }}_{U_i\cap U_j}⟶{\mathcal{O}}_X{\mathrm{\mid }}_{U_i\cap U_j},s⟼g_{ij}s\mathrm{.}$ Because $g_{ij}$ is a unit, ${\phi }_{ij}$ is an isomorphism of ${\mathcal{O}}_X$-modules, and the rules for the $g_{ij}$ are exactly the identity and cocycle conditions of [F2].

By [L1], the local free rank-one modules ${\mathcal{O}}_X{\mathrm{\mid }}_{U_i}$ glue to an ${\mathcal{O}}_X$-module $\mathcal{L}$ on $X$ with $\mathcal{L}{\mathrm{\mid }}_{U_i}\cong {\mathcal{O}}_X{\mathrm{\mid }}_{U_i}$ for every $i$. Therefore $\mathcal{L}$ is locally free of rank one, i.e. a line bundle.

Let $\mathcal{P}$ be the presheaf cokernel of the inclusion $\underset{‾}{\mathbb{Z}}\hookrightarrow \mathcal{C}$, so $\mathcal{P}(V)=\mathcal{C}(V)\mathrm{/}\underset{‾}{\mathbb{Z}}(V)$ for each open $V\subseteq S^1$. Cover $S^1$ by the arcs $U_0=S^1\setminus \{(-1,0)\}$ and $U_1=S^1\setminus \{(1,0)\}$, and choose continuous angle functions ${\theta }_0:U_0\to (-1\mathrm{/}2,1\mathrm{/}2)$ and ${\theta }_1:U_1\to (0,1)$ with $e^{2\pi i{\theta }_i(z)}=z$. On $U_0\cap U_1$, the difference ${\theta }_1-{\theta }_0$ is locally constant integer-valued, so the classes $[{\theta }_0]\in \mathcal{P}(U_0)$ and $[{\theta }_1]\in \mathcal{P}(U_1)$ agree on the overlap.

If these local classes came from a global class $[g]\in \mathcal{P}(S^1)$, then on each arc $U_i$ the difference $g-{\theta }_i$ would be integer-valued and continuous, hence constant because $U_i$ is connected. On the two connected components of $U_0\cap U_1$, this would force the same constant difference to be both $0$ and $1$, which is impossible. Therefore the compatible local classes of step 1.1 do not glue in the presheaf $\mathcal{P}$.

So the objectwise cokernel presheaf is not a sheaf. By [F1], this is exactly why one sheafifies it to obtain the true cokernel sheaf. The statement is false.