Defining subobjects

Many of the notions we have in familiar categories, like products, coproducts and injective/surjective homomorphisms, can be generalized straightforwardly to a general categorical setting. For instance, the direct product of groups in $\mathbf{Grp}$, the topological product in $\mathbf{Top}$, and the greatest lower bound in any poset are just specific instances of the general categorical product.

However, I have come to realize that not all notions are equally easy to generalize to a categorical setting, and one of them is the notion of subobjects. In algebra, we are used to subgroups, subrings, submodules and so on. In more topological categories like $\mathbf{Top}$ and the category of metric spaces and decreasing maps, every subset inherits a subspace topology and subspace metric, respectively. Ideally, all of these notions could be generalized to a categorical notion of subobjects.

What is common among all these concrete categories is that the inclusion of a subobject into an object is injective as a map between sets, and also a homomorphism with respect to the given structure, whether it's an algebraic operation or a topology. As morphisms in concrete categories with injective underlying functions are always monomorphisms (exercise!), one might think that the correct notion of a subobject of an object $X$ in a category $\mathcal{C}$ is an object $A$ together with a monomorpism $A \rightarrowtail X$.

The first problem one realizes with this definition is that we get too many subobjects. For instance, two monomorphisms into an abelian group could represent the same subgroup. Take for instance the abelian group $\mathbb{Z}$. Here, both the inclusion map $2 \mathbb{Z} \hookrightarrow \mathbb{Z}$ and the multiplication by 2 map $\cdot 2 : \mathbb{Z} \to \mathbb{Z}$ represent the same subgroup, as the image of $\cdot 2$ is precisely $2\mathbb{Z}$. The observation here is that two subobjects $A \rightarrowtail X$ and $B \rightarrowtail X$ should be considered equivalent if there exists an isomorphism $\phi : A \to B$ such that the diagram

$$\begin{matrix} A & \overset{\phi}{\longrightarrow} & B \\ & \searrow & \downarrow \\ & & X \end{matrix}$$

commutes. It can easily be verified that this is an equivalence relation, although we have to be careful here. The class of monomorpisms into an object $X$ is in general not a set. However, in many of our usual categories, it is a set up to equivalence: That is, a class containing a representative of each monomorphism into $X$ up to equivalence will be a set. Such categories are called well-powered, and we won't deal with other categories in this post. If $\mathcal{C}$ is well-powered, we denote by $\text{Sub}_{\mathcal{C}}(X)$ the set of (equivalence classes) of subobjects of $X$.

Now, the definition above turns out to be satisfactory in many algebraic categories. I will give a proof in the case of $\mathbf{Grp}$, which also holds in many other categories.

Proposition 1. In the category of groups, actual subgroups of a given group $G$ are in one-to-one correspondence with $\text{Sub}_{\mathbf{Grp}}(G)$.

Proof: We define a mapping from $\text{Sub}_{\mathbf{Grp}}(G)$ to $\{ \text{subgroups of $G$} \}$ as follows: If $f : A \rightarrowtail G$ is a monomorphism into $G$, we associate to it the subgroup $\text{Im}(f)$ of $G$. This mapping is surjective, as if $G' \subseteq G$ is a subgroup, then the inclusion $G' \hookrightarrow G$ is a monomorphism. Now if two monomorphisms $f : A \rightarrowtail G$ and $g : B \rightarrowtail G$ into $G$ are equivalent, then $\text{Im}(f) = \text{Im}(g)$. Conversely, if $\text{Im}(f) = \text{Im}(g)$, then the map $A \rightarrow \text{Im}(f) = \text{Im}(g) \to B$ is an isomorphism. This proves that the mapping is a bijection.  $\square$

Essential to the proof above is that if $f: G \to G'$ is a bijective homomorphism, then it is an isomorphism, i.e. its inverse is also a group homomorphism. We used this when we inverted a monic from its image. As we know, this property does not hold in $\mathbf{Top}$, for instance. That is, a continuous bijection can have a discontinuous inverse. Indeed, the following example shows that there are more monics into a topological space $X$ up to equivalence than there are subspaces of $X$.

Example 1. Let $f: [0,1) \to \mathbb{C}$ be given by $f(t) = e^{2\pi i t}$. This is an injective continuous map, hence a monomorphism in $\mathbf{Top}$. Assume that $f$ is equivalent to an actual subspace of $X$, i.e. there exists a homeomorphism $\phi : [0,1) \to A$, where $\iota : A \hookrightarrow X$, such that $f = \iota \circ \phi$. Then $A=\text{Im}(\iota) = \text{Im}(f) = S^1$, the circle viewed as a subset of the complex plane. Now pick a small connected neighbourhood $U$ of the point $(1,0)$. Then the preimage $f^{-1}(U)$ consists of two disjoint open sets in $[0,1)$. But we also have that
$$f^{-1}(U) = (\iota \circ \phi)^{-1}(U) = \phi^{-1}(U)$$
Thus the preimage of a connected set under $\phi$ is disconnected, which is a contradiction as $\phi$ is a homeomorphism. Hence, $f$ is not equivalent to any actual subspace of $\mathcal{C}$. $\triangle$

The moral here is that monomorphisms can be too general to correspond to the preferred notion of subobject. In fact, it can be showed that if we restrict to regular monomorphisms, i.e. equalizers of pairs of morphisms (which are always monic), then $\text{Sub}_{\mathbf{Top}}(X)$ would be in one-to-one correspondence with the actual subspaces of $X$.

One approach is to speak of $M$-subobjects instead of just subobjects, where $M$ is some class of special monomorphisms into $X$. For instance, they could be regular monomorphisms, strong monomorphisms, split monomorphisms, and so on. In $\mathbf{Ab}$, one would want $M$ to contain all monomorphisms, while in $\mathbf{Top}$, one would only want the regular ones. Interestingly, if one only wants normal subgroups in $\mathbf{Grp}$, one would restrict to so-called normal monomorphisms.

But still, what can we conclude from these observation? If one really thinks of category theory as the correct way to look at these structures, then maybe the definitions of subobjects we have in various categories are somewhat arbitrary? Why should a subset of a topological space be equipped with the subspace topology and not some topology that merely makes the inclusion map a monomorphism and not something stronger? I find these philosophical questions interesting. Personally, I think that the category structure here is not enough, and that one should considered these categories as carrying more structure. Only then will it be possible to make sense of why we want the subspace topology instead of other topologies, in an (almost) categorical sense.

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.