What are monomorphisms in the category of real vector bundles over a fixed base space $X$?

Question

I have a (perhaps stupid) question dealing with vector bundles.

In most textbooks, a morphism of real vector bundles $f:\xi\to \eta$ over a fixed base space $X$ is said to be a $\textsf{monomorphism}$ if the total space map $f:E\to E^{\prime}$ is (fiberwise) injective. (Epimorphisms are analogously defined fiberwisely.)

What I want to check is that this definition matches with the definition of $\textsf{monomorphism}$s in the category of vector bundles over $X$ (let us denote this as $\textbf{VB}_{X}$), namely a morphism which can be cancelled from left. Vector bundle morphisms with injective total space maps are obviously monomorphisms in this category-theoretic sense. However, I can't show the converse, that $\textbf{VB}_{X}$-monomorphisms necessarily have injective total space maps (I can't find a thing like an "indicator sheaf" for sheaves in this vector bundle case...). I now even suspect that these two notions are not equivalent, but I can't find a good counterexample. Any helps would be appreciated!

Answer 1

$\text{Vect}(X)$ isn't abelian, but it is still close enough that a morphism of vector bundles is a monomorphism iff it has trivial kernel. The problem is that the fiberwise kernel of a map of vector bundles can fail to be a vector bundle.

Very explicitly, let $X = \mathbb{R}$, let $V$ be the trivial line bundle over $X$, and let $f : V \to V$ be the morphism which is given fiberwise by

$$f(v_x) = x v_x, x \in \mathbb{R}, v_x \in V_x.$$

This map fails to be fiberwise injective at the origin, and in fact its fiberwise kernel jumps in dimension there. I claim it is a monomorphism in vector bundles. The reason is that if $W$ is another vector bundle and $g : W \to V$ a map of vector bundles such that $f \circ g = 0$, then since $f$ is fiberwise injective everywhere except the origin, $g$ must be zero everywhere except the origin, and then by continuity it must be zero at the origin as well.

As the comments indicate, to make things work out better you can try to embed vector bundles into a larger and nicer category (abelian would be nice). It needs to be at least large enough that the fiberwise kernel of the above map (one description of which is as a skyscraper sheaf at the origin) exists in it. If $X$ is compact Hausdorff, then by the Serre-Swan theorem $\text{Vect}(X)$ is equivalent to the category of finitely generated projective modules over the ring $C(X)$ of continuous functions $X \to \mathbb{R}$, and this naturally embeds into the category of all finitely generated modules, which in turn embeds into the category of all modules, and that's certainly abelian. From here you can pick out the vector bundles as the dualizable objects, or as the tiny objects. In general, e.g. in algebraic geometry, you can instead work with various categories of sheaves of $\mathcal{O}_X$-modules.

Answer 2

Here is an algebraic version (and generalization) of Qiaochu's counterexample:

Let $A$ be a commutative ring and $a : A \to A$ be a homomorphism of $A$-modules (free of rank $1$), corresponding to some element $a \in A$.

It is a monomorphism in the category of f.g. projective $A$-modules iff it is a monomorphism in the category of all $A$-modules iff $a$ is regular, i.e. no zero-divisor.

It is a monomorphism in each fiber iff for all prime ideals $\mathfrak{p}$ the element $\overline{a} \in \mathrm{Frac}(A/\mathfrak{p})$ is regular, i.e. non-zero. This means that $a$ is not contained in any prime ideal.

The set of zero divisors of $A$ is a union of prime ideals. This shows (what we already know for other reasons) that fiberwise-mono implies mono. But since the set of zero divisors does not have to be the union of all prime ideals, there are monos which are not fiberwise-mono:

In the ring $A=C(\mathbb{R})$, the function $a$ defined by $a(t)=t$ is no zero divisor (because if $f(t) \cdot t = 0$ for all $t$, this implies $f(t)=0$ for all $t \neq 0$ and hence $f=0$ since $\mathbb{R} \setminus \{0\}$ is dense in $\mathbb{R}$), but $a$ is contained in the prime ideal of functions which vanish at $0$.