TITLE: What are monomorphisms in the category of real vector bundles over a fixed base space $X$?
QUESTION [8 upvotes]: 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!

REPLY [4 votes]: 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 $0$.