Is an injective function always monomorphic

Question

Is an injective function monomorphic (in category of sets)? How do I prove this?

Conversely, I could prove that a monomorphic function is injective in the following manner:

If f is monic, then $f\circ g = f\circ h\Rightarrow g= h$

Given the mapping $C\to A\to B$, where $g,h\colon C\to A$ and $f\colon A\to B$ Assume $f$ is non-injective then there are some $a$, $b$ in $A$ and $c$ in $B$ such that $f(a) = f(b) = c$

There are also $x$ and $y$ in $C$ such that $g(x) = a$ and $h(y) = b$ Thus $f(g(x) = f(h(y)) = c$, thus $f\circ g = f\circ h$, but $g \neq h$ but $g = h$ Thus $f$ cannot be non-injective

I could not prove it other way round, so is an injective function monic?

Answer 1

Suppose $f\colon A\to B$ is monic. Let $x,y\in A$ and consider the maps \begin{align} g&\colon \{x,y\}\to A, &&g(x)=g(y)=x\\ h&\colon \{x,y\}\to A, &&h(x)=h(y)=y \end{align} If $f(x)=f(y)$, then $f\circ g=f\circ h$. Since $f$ is monic, we derive $g=h$ and so $x=y$. Therefore $f$ is injective.

Assume $f\colon A\to B$ is injective and that $g,h\colon C\to A$ satisfy $f\circ g=f\circ h$. Then, for every $x\in C$, $$ f(g(x))=f(h(x)) $$ and, by the injectivity of $f$, $g(x)=h(x)$. Therefore, $g=h$ and $f$ is monic.

Note that the second argument works in every concrete category.

Answer 2

Injections split$^1$ in $\text{Set}$. A splitting morphism is always a mono.

1. There is a map $B \to A$ such that $A \stackrel{f}{\to} B \to A$ is the identity of $A$.

Answer 3

In fact, in any concrete category over $\mathbf{Set}$, injections are monomorphisms.

To see this, let $(A, U)$ be such a category with associated forgetful (faithful) functor. Then if $f$ is such that $Uf$ is an injection, and if $f\circ g = f\circ h$, then $Uf \circ Ug = Uf\circ Uh$

So to prove this fact it suffices to prove it for $\mathbf{Set}$, which other answers have already done: if $f$ is injective in $\mathbf{Set}$, and $f\circ g = f\circ h$, then for all $x$, $f(g(x))=f(h(x))$, so by injectivity, $g(x)= h(x)$, and so $g=h$. Therefore we get $Ug = Uh$, and since $U$ is faithful and the codomains/domains are the right ones, we can conclude $g=h$: $f$ is monic.

What's interesting is that the converse is sometimes not true; but it is whenever $U$ is representable, which happens for instance when $U$ has a left adjoint. Indeed, representable functors preserve monos, so if $f$ is mono and $U$ representable, then $Uf$ is mono, and therefore injective. That's why in categories such as $\mathbf{Grp}$ (or any "algebraic category"), $\mathbf{Top}$, mono means the same as injective (for these categories, $U$ has a left adjoint and so is representable, because $Ux \cong Hom(1, Ux) \cong Hom(F1, x)$, so $U\cong Hom(F1, -)$)