Image factorizations and epimorphisms

We are used to taking images of morphisms in many concrete categories. That is, given a morphism $f: A \to B$ in a category $\mathcal{A}$, we often form the subset
$$\text{Im}(f) = \{ f(a) : a \in A \}$$
of the underlying set of $B$. This turns out to be an object in its own right in many familiar algebraic categories, and we can always factor $f$ through the image as $A \rightarrow \text{Im}(f) \hookrightarrow B$.

A standard categorical formulation of images is as follows: Given a morphism $f: A \to B$ in $\mathcal{A}$, an image of $f$ consists of a factorization $A \overset{g}{\rightarrow} I \overset{\iota}{\rightarrowtail} B$ where $\iota$ is monic, such that for any other factorization $A \to C \rightarrowtail B$ with the last morphism monic, there exists a unique morphism $I \to C$ making the triangles in the following diagram commute:
$$\begin{matrix} A & \overset{g}{\rightarrow} & I \\ \downarrow & \swarrow & \downarrow \small \iota \normalsize \\ C & \rightarrow & B \end{matrix}$$
Because this is a universal property, this characterizes $(I, g, \iota)$ up to unique isomorphism. Thus, if it exists we denote $I$ by $\text{Im}(f)$ and call it the image of $f$.

In most of the categories we are used to taking images in, the morphism $g: A \to \text{Im}(f)$ is always an epi, although this is not stated in the categorical definition. The question is: Does it follow from the definition that $g$ is epi, or do there exist examples of categories where it isn't?

The answer turns out to be the latter. The main goal of this blog post, however, is to show that under the assumption of existence of some limits, $g$ is indeed epic. The main proposition of the blog post is the following:

Proposition. Suppose $\mathcal{A}$ has pullbacks and equalizers. Then if $f: A \to B$ admits an image factorization $A \overset{g}{\to} \text{Im}(f) \overset{\iota}{\rightarrowtail} B$, the morphism $g$ is epic.

We will prove this proposition by a series of three lemmas. First, define a morphism $f: A \to B$ to be an extremal epimorphism if whenever $f = m \circ g$ where $m$ is a mono, then $m$ is an isomorphism. Note that I haven't required extremal epimorphisms to be epimorphisms. In fact, they need not be.

Lemma 1. If $f: A \to B$ has image factorization $f = \iota \circ g$, then $e$ is an extremal epimorphism.

Proof: Suppose $g = m \circ h$, where $m$ is monic. Then $f = \iota \circ (m \circ h) = (\iota \circ m) \circ h$, and so we have found a factorization of $f$ by a morphism $h$ followed by a mono $\iota \circ m$. By the universal property of images, there exists a morphism $j$ such that $(\iota \circ m) \circ j = \iota$. Since $\iota$ is monic, this gives $m \circ j = \text{id}$. As $m$ is monic and has a right inverse, it is an isomorphism. $\square$

Let $f : A \to B$ and $g: C \to D$ be morphisms. We say that $f$ is orthogonal to $g$ if for any commutative square
$$\begin{matrix} A & \overset{f}{\rightarrow} & B \\ \downarrow  &  & \downarrow \\ C & \overset{g}{\rightarrow} & D \end{matrix}$$
there exists a  unique morphism $k: B \to C$ making both the triangles commute:
$$\begin{matrix} A & \overset{f}{\rightarrow} & B \\ \downarrow  & \overset{k}{\swarrow}  & \downarrow \\ C & \overset{g}{\rightarrow} & D \end{matrix}$$

Observe that if $f$ is orthogonal to a monomorphism $g$, then the $k$ making the appropriate triangle commute is automatically unique. In other words, when proving that $f$ is orthogonal to some monic, it is enough to show existence of $k$.

Lemma 2. Let $f: A \to B$ be an extremal epimorphism. Then $f$ is orthogonal to all monomorphisms.

Proof: Suppose
$$\begin{matrix} A & \overset{f}{\longrightarrow} & B \\ \small h \normalsize \downarrow & & \downarrow \small  j \normalsize \\ C & \overset{m}{\longrightarrow} & D \end{matrix}$$
is a commutative square. Also, let
$$\begin{matrix} P & \overset{n}{\longrightarrow} & B \\ \small r \normalsize \downarrow & & \downarrow \small  j \normalsize \\ C & \overset{m}{\longrightarrow} & D \end{matrix}$$
be a pullback square. Then by the universal property of pullbacks, there exists a morphism $\alpha : A \to P$ such that $f = n \circ \alpha$. Now by a general property of pullbacks, $n$ is monic since $m$ is monic. By Lemma 1, $f$ is an extremal epimorphism, so we conclude that $n$ is an isomorphism. Let $\phi : B \to P$ be its inverse. Then if $k$ is the composition $B \overset{\phi}{\to} P \overset{r}{\to} C$, both the triangles
$$\begin{matrix} A & \overset{f}{\longrightarrow} & B \\ \small h \normalsize \downarrow & \overset{k}{\swarrow} & \downarrow \small  j \normalsize \\ C & \overset{m}{\longrightarrow} & D \end{matrix}$$
commute. It is also unique, as $m$ is monic. This finishes the proof. $\square$.

Combining the two lemmas, we have shown that $g$ in an image factorization $f = \iota \circ g$ of $f$ is orthogonal to all monomorphisms. We finish the proof of our proposition by showing that if equalizers exist, then a morphism orthogonal to all monomorphisms must be an epimorphism.

Lemma 3. If $\mathcal{A}$ has equalizers and $f: A \to B$ is orthogonal to all monomorpisms in $\mathcal{A}$, then $f$ is an epimorphism.

Proof: Suppose we have morphisms
$$A \overset{f}{\to} B \underset{h}{\overset{g}{\rightrightarrows}} C$$
such that $g \circ f = h \circ f$. Form the equalizer $E \overset{m}{\to} B$ of the parallel arrows $B \underset{h}{\overset{g}{\rightrightarrows}} C$. By the universal property of equalizers, there exists a unique $j : A \to E$ such $m \circ j = f$.
$$\begin{matrix} A & \overset{f}{\longrightarrow} & B & \underset{h}{\overset{g}{\rightrightarrows}} & C \\ \small j \normalsize \downarrow & \overset{m}{\nearrow} & & & \\ E & & & \end{matrix}$$
It is a property of equalizers that the morphism $m$ is monic. By orthogonality, there exists $k : B \to E$ such that the triangles of the commutative square
$$\begin{matrix} A & \overset{f}{\longrightarrow} & B \\ \small j \normalsize \downarrow & \overset{k}{\swarrow} & \downarrow \small \text{id}_B \normalsize \\ E & \overset{m}{\longrightarrow} & B \end{matrix}$$
commute. That is, $m \circ k = \text{id}_B$, making $m$ a mono with a right inverse, i.e. an isomorphism. Thus, since we have $g \circ m = h \circ m$, we get $g = h$ by cancellation. This proves that $f$ is an epimorphism. $\square$.

It is a well known fact that existence of products and equalizers imply existence of pullbacks, so one could replace the assumption of pullbacks by the assumption of products in the Proposition, if one wants to.

So far, we have discussed when $g$ in an image factorization $f = \iota \circ g$ is epic, given that the image factorization already exists. But when does an image factorization of a morphism $f$ exist, in general? It turns out that if $\mathcal{A}$ satisfies the following properties:

• $\mathcal{A}$ has pushouts and equalizers.
• Every monomorphism $m : A \to B$ in $\mathcal{A}$ is a regular monomorphism: That is, $m : A \to B$ is isomorphic to the equalizer of some pair $B \underset{f}{\overset{g}{\rightrightarrows}} C$.

Then image factorizations of every morphism exist (see this entry at nLab). Do our favorite categories satisfy the above properties? Well, most of our favorite categories have finite limits and colimits, so the first property is satisfied. The second one is also often satisfied, but is a bit more tricky to prove.

It's not so bad in $R\mathbf{Mod}$: We can take $C$ to be $B/\text{Im}(f)$, $f$ to be the quotient map and $g=0$. One can generalize this straightforwardly to any Abelian category. It's harder to prove the statement in the category of groups, see this link.