Are monomorphisms in all concretizable categories with certain objects injections? Specifically, I'm thinking of a concretizable category $C$, whose objects are sets with extra structure and whose arrows are set-functions satisfying certain conditions. Suppose that there is a faithful functor $U:C\rightarrow \textbf{Set}$, which sends $C$-objects to their underlying sets and sends $C$-arrows to their corresponding functions (the same functions set-theoretically speaking).
Suppose also that there is a $C$-object which is mapped to the a singleton in $\textbf{Set}$ by $U$. Since a singleton is contained in the image of $C$ under $U$: $img(C)$ (suppose that it is a category), one can show that monomorphisms in the $img(C)$ are injections. Since faithful functors reflect monomorphisms, a $img(C)$-arrow $U(f)$ being monic implies that the $C$-arrow $f$ is also monic. Since $f$ and $U(f)$ are the same function and that monomorphisms in $img(C)$ are injective, does this mean that monomorphisms in $C$ are injections?
 A: The standard example of a concrete category (of algebras, in the sense of universal algebra) where not every monomorphism is injective is the category of divisible abelian groups, with group morphisms between them. In this category, the projection map $\mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$ is a monomorphism (see for example the proof in Wikipedia).
Note that this category includes an object on a one-element set. Your error is thinking that this allows you to prove monomorphisms are injective: the functions in $\mathsf{Set}$ that you would use to prove injectivity have no counterpart in this category, since the trivial group is not free, but is instead a zero object. The concretization functor is required to be faithful; it is not required to be full. You are implicitly assuming it is full in that "one can show".
What is true is that if you have a "free object" in a one element set, then monomorphisms will be injective. By "free object on a set $X$" I mean the image of $X$ under a left adjoint to the concretization functor. Note that there is no "free divisible group" on a non-empty set, and likewise in Qiaochu's example, there is no free object in one element.
A: No. Here is the sort of thing that can go wrong: take $C = \text{Ab}$ to be the category of abelian groups, but take the forgetful functor $U : C \to \text{Set}$ to be
$$U : A \mapsto A \oplus A/2A.$$
Then, for example, multiplication by $2$ is a monomorphism $\mathbb{Z}/2 \to \mathbb{Z}/4$ but it doesn't remain injective after applying $U$.
What is true is that if $U$ preserves pullbacks then it preserves monomorphisms, because a morphism $f$ is a monomorphism iff its kernel pair (which is the pullback of $f$ against itself) is trivial. So in particular this is the case if $U$ is representable, which is the nicest case; nearly all standard forgetful functors for categories of algebraic objects arise in this way since the representing object is typically the free object on one generator, e.g. for the standard forgetful functor for $\text{Ab}$ the representing object is $\mathbb{Z}$.
