In Bredon's Sheaf Theory textbook, he introduces the notion of a subsheaf on page 9; here $X$ is a topological space.
Definition 2.1: A "subsheaf" $A$ of a sheaf $B$ is an open subspace of $B$ such that $A_x=A\cap B_x$ is a subgroup of $B_x$ for all $x\in X$. (That is, $A$ is a subspace of $B$ that is a sheaf on $X$ with the induced algebraic structure.)
My questions are:
1.Do we require $A\subset B$ to be open or not? He says open at first then omits open later.
- If we do require $A\subset B$ to be open, how can the kernel of a sheaf homomorphism be a subsheaf? For example, a vector bundle is a sheaf, but the kernel of bundle homomorphism is closed, not open. We have a similar problem with the image of a vector bundle homomorphism.
Thanks in advance!
Yes, a subsheaf is required to be an open subspace. This is actually automatic from saying that $A$ itself is a sheaf over $X$, though. For $A$ to be a sheaf, the projection map $p:A\to X$ must be a local homeomorphism, so for any $a\in A$, there is an open subset $U\subseteq A$ containing $a$ such that the restriction of $p$ to $U$ is a homeomorphism to an open subset of $X$. Since $B$ is a sheaf, there is also an open subset $V\subseteq B$ containing $a$ such that the restriction of $p$ to $V$ is a homeomorphism to an open subset of $X$. Then we may replace $U$ with $U\cap V$ and assume that $U$ is contained in $V$. But now we see that $U$ is open in $V$, since $p(U)$ is open in $X$ and hence in $p(V)$ and $p$ restricts to a homeomorphism $V\to p(V)$. Since $V$ is open in $B$, this implies $U$ is open in $B$, so $A$ contains an open neighborhood of $a$ in $B$. Since $a\in A$ was arbitrary, this means $A$ is open in $B$.
As for your question about kernels, keep in mind that a vector bundle (as a space over $X$ whose fibers are vector spaces) is not a sheaf over $X$. Rather, it has an associated sheaf whose points are germs of sections of the vector bundle (which in some sense contains the same data as the vector bundle and so in this way you can think of vector bundles as sheaves). If you have a homomorphism $f:E\to F$ of vector bundles over $X$ then the kernel of the corresponding homomorphism of sheaves consists of those germs of sections of $E$ that become the $0$ germ after composition with $f$. That means that in some neighborhood of the point, they become identically $0$ after composition with $f$. Those neighborhoods correspond to neighborhoods in the sheaf of germs of sections of $E$ which show the kernel is an open subset.
More generally, if $B$ is a sheaf of groups on $X$ then the set of identity elements of stalks of $B$ is an open subset of $B$ (this follows from the first paragraph, since the set of identity elements is a subsheaf of $B$, being the image of a sheaf map $X\to B$). So the kernel of a homomorphism of sheaves of groups $A\to B$ is open, being the inverse image of this open subset.