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.

Subrngs and subrings of rngs and rings

In this post, I will discuss what different notions of subrngs and subrings in rngs and rings.

I will start by discussing rngs. I define a rng as follows: A rng is a tuple $(R,+,0,i,\cdot)$ where

• R is a set
• $+ : R \times R \to R$ is a map called addition. We write $+(a,b) = a+b$.
• $0 \in R$ is an element called the additive identity.
• $i : R \to R$ is a map called additive inversion, and we write $i(a) = -a$.
• $\cdot : R \times R \to R$ is a map called multiplication. We write $\cdot(a,b) = a \cdot b$

The following axioms must be satisfied: For all $a,b,c \in R$, we need:

• $a+(b+c) = (a+b)+c$.
• $a \cdot (b \cdot c) = (a \cdot b) \cdot c$.
• $a+b=b+a$.
• $a+0=a$.
• $a+(-a) = 0$.

From here on, I will refer to a typical rng $(R,+,0,i,\cdot)$ by $R$, where the rest of the structure is understood. We will then denote the structure of $R$ by $+_R$, $0_R$, and so on.

A rng morphism from R to S is a function $f: R \to S$ that respects all the rng structure. That is, the following properties must be satisfied for all $a,b \in R$:

• $f(a+_Rb) = f(a)+_Sf(b)$.
• $f(0_R)=0_S$.
• $f(i_R(a))=i_S(f(a))$.
• $f(a \cdot_R b) = f(a) \cdot_S f(b)$.

As you might know, it turns out that the preservation of addition property implies preservation of the additive identity and additive inversion, so it is enough to check the first and last properties. Still, we include all four as axioms, because these implications are not a priori obvious.

The collection of rngs and rng morphisms form the category Rng of rngs.

So what is the proper notion of a subrng of a rng $R$? It should be a subset $S$ of the set $R$ that is a rng in itself with the inherited structure from $R$. Thus we need closure under addition and multiplication as well as closure under inversion and closure under the additive identity, which is to say that $0_R \in S$. All this can be stated alternatively by requiring that whatever rng structure we put on $S$, we want the inclusion map $S \to R$ to be a rng morphism (a requirement that uniquely specifies the rng structure of $S$, if it is possible to equip it with such a compatible rng structure at all).

Now what is a ring? I can think of two sensible definitions. The first one is as follows: A ring is a rng $R$ such that there exists some element $1 \in R$ with $a \cdot_R 1 = 1 \cdot_R a = a$ for all $a \in R$. It turns out that such an element is unique if it exists. With this definition, a ring is just a rng that happens to satisfy some extra property, just like other properties a ring can satisfy, e.g. being commutative, being artinian, and so on. Defining rings in this way, we can think of the category Ring of rings as a full subcategory of Rng.

Here's the other definition: A ring is a tuple $(R,+,0,i,\cdot,1)$ where $(R,+,0,i,\cdot)$ is a rng (we can spell out all the axioms again if we want to) and $1$ is an element of $R$ such that for all $a \in R$, we have that $a \cdot 1 = 1 \cdot a = a$ for all $a \in R$. What is the difference between this definition and the previous definition, you might ask. The two definitions might seem identical at first, but in my view they are quite different. In this definition, a ring isn't just a special type of rng, it is rather a rng equipped with extra structure, just like a vector space is an abelian group equipped with extra structure. This is a crucial difference, because when we have extra structure, it is natural to require the morphisms to preserve this extra structure: The right notion of a ring morphism is thus a map $f: R \to S$ that is a rng morphism of the underlying rngs, but which also preserves the multiplicative identities: That is, $f(1_R) = 1_S$. Hence, with this definition, the category Ring of rings cannot be viewed as a full subcategory of Rng, as the morphisms are more specific. We still have a forgetful functor from Ring to Rng that sends a ring to its underlying rng.

Let's examine how the two different definitions affect the notions of subrings of rings. Let $R$ be a ring. With the first definition, $R$ is just a rng that happens to contain a multiplicative identity. A subring $S$ of $R$ is then a subrng of $R$ which also happens to contain a multiplicative identity. This multiplicative identity might not be the same multiplicative identity as in $R$! Take for instance the rng $R=\mathbb{Z}_2 \times \mathbb{Z}_2$ with the subrng $\{ 0 \} \times \mathbb{Z}_2$. The rng $R$ is a ring, as $(1,1)$ is a multiplicative identity. However, $S$ is also a ring as $(0,1)$ is a multiplicative identity in $S$.

This could never happen with the second definition. Here, the multiplicative identity $1$ is already part of the structure of the ring $R$. A subring $S$ of $R$ would have to inherit all the structure from $R$, including its multiplicative identity: Otherwise, the inclusion map $S \to R$ wouldn't map $1_S$ to $1_R$, and then it wouldn't be a ring homomorphism! We could still talk of subrngs of rings. This would then be a subset that satisfied all the properties of a subring except possibly containing the identity of $R$. Thus, a subrng containing an identity different from the identity of $R$ would be considered a subrng, but not a subring.

To summarize, we have the four following concepts:

• Subrng $S$ of rng $R$
• Subrng $S$ of ring $R$
• Subring $S$ of rng $R$
• Subring $S$ of ring $R$

The first one is unambiguous, as we have only one definition of rng. The second one is also unambiguous, because it gives us the same concept regardless of which definition of ring we use. The third one is unambigious, because $R$ doesn't have an identity to begin with. The fourth one however, is ambiguous, as in the first definition, a subring is allowed to have an identity different from the identity of $R$, while in the second, the identity must be the same in both rings.