A monomorphism that is not injective

I am trying to find an example of a monomorphism that is not an injective map; much like there exist epimorphisms that are not surjective. Is this a bad question? Is every monomorphism defined on the category of sets injective? Any help would be great.

Solution 1:

In the category of divisible (abelian) groups, the map $\pi\colon\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$ is a non-injective monomorphism.

Indeed, suppose that $f,g\colon G\to \mathbb{Q}$ are such that $\pi\circ f= \pi\circ g$. Let $x\in G$. Then $f(x)-g(x) = n\in\mathbb{Z}$. If $n\neq 0$, then let $y\in G$ be such that $y^{2n}=x$ (using multiplicative notation for $G$ since we are not assuming $G$ is abelian). Then $f(x) = f(y^{2n}) = 2nf(y)$, hence $f(y) = \frac{1}{2n}f(x)$; and similarly $g(y) = \frac{1}{2n}g(x)$. Now, $f(y)-g(y)$ must be an integer, but $$f(y)-g(y) = \frac{1}{2n}(f(x)-g(x)) = \frac{1}{2},$$ a contradiction. Therefore, $n=0$, so $f(x)=g(x)$. Thus, $f=g$ and $\pi$ is a monomorphism.

One reason why non-injective monomorphisms are hard to find in algebraic contexts is that the existence of a free object on the one element set (in concrete categories) implies that monomorphisms are one-to-one, and in most "familiar" categories of algebras (in the sense of general algebra), free objects exist.

To see the implication above, let $F_x$ be the free object on the set $\{x\}$, and let $m\colon A\to B$ be a monomorphism. If $m(a) = m(a')$, then the maps $f,g\colon F_x\to A$ induced by $x\mapsto a$ and $x\mapsto a'$, respectively, satisfy $m\circ f = m\circ g$, and hence $f=g$, thus $a=a'$, proving that $m$ is one-to-one.

Solution 2:

Consider the category of pointed, connected, locally connected and locally path-connected spaces. Any nontrivial covering map is a monomorphism in this category which is not injective on underlying sets; this is a restatement of one of the lifting properties of covering maps.