Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The book I'm working with (Mitchell's Theory of Categories (1965)) defines a $A'$ to be a subobject of an object $A$ (in some category $\mathbf{C}$ ) iff there exists some monomorphism $\alpha\!:\!A' \rightarrowtail A$. (I quote Mitchell's exact definition along with one important caveat in another question on math.SE). In addition, Mitchell calls such $\alpha$ the inclusion of $A'$ in $A$, and defines the notation $A' \subset A$ to mean $A'$ is a subobject of $A$ . Furthermore, he adds that in this case "we shall say that $A'$ is contained in $A$, or that $A$ contains $A'$".

Mitchell's choices of notation and nomenclature strongly suggest that he intends the concept of subobject to be a generalizations of the concept of subset from standard set theory. If this is indeed the case, I find the whole idea really puzzling, because the analogy with subsets seems to me more likely to be confusing than illuminating.

What is the benefit of pushing this analogy between subsets and subobjects? Is there a way to modify Mitchell's definitions that retain this benefit without the flaws described below?

I find that, when working with subobjects as Mitchell defines them, my intuition frequently trips on one or the other of two important differences between the is-subset-of and is-subobject-of relations, as described below; they look like significant flaws to me.

First, in set theory $(A \subset B) \wedge (B \subset A)$ is equivalent to $A = B$, but this equivalence does not hold with Mitchell's $\subset$. Certainly, $A = B$ implies that $(A \subset B) \wedge (B \subset A)$ (in Mitchell's sense), but the reverse implication does not hold. Indeed Mitchell's $(A \subset B) \wedge (B \subset A)$ ensures only the existence of monomorphisms $\alpha\!:\!A \rightarrowtail B$ and $\beta\!:\!B \rightarrowtail A$. (In fact, as far as I can tell, one cannot even say that such $A$ and $B$ are isomorphic!)

Second, suppose that $A$ is a subobject of $B$ and $B$ is a subobject of $C$, with inclusions* $\alpha\!:\!A\rightarrowtail B$ and $\beta\!:\!B\rightarrowtail C$. In this context, the statement $A$ is a subobject of $C$ admits two inequivalent interpretations:

  1. it simply asserts a special case of the transitivity of the is-subobject-of relation (which follows from the fact that the composition of monomorphisms is also a monomorphism);
  2. it asserts that there exists a monomorphism $\gamma\!:\!A\rightarrowtail C$ (irrespective of the existence of $\beta\;\alpha$).

Clearly, the first interpretation implies the second one, but the converse is not true.

Contrast this situation with the one in which $A$ is a subset of $B$ and $B$ is a subset of $C$. In this case, the statement $A$ is a subset of $C$ can be interpreted only as a special case of the transitivity of the is-subset-of relation. There is no alternative way in which $A$ can be a subset of $C$ (given that $A$ is a subset of $B$ and $B$ is a subset of $C$).

Actually, Mitchell's book deals only with categories having the property that for every pair $(A, B)$ of category objects, the class of all morphisms $A\to B$ is a set. The Wikipedia page for category theory calls such categories locally small, but I'm refraining from using this term here because Mitchell defines locally small categories differently. The definition chasing never ends...

In my experience, in the context of standard set theory, $A \subset B$ usually implies that $A \neq B$. This is not the case with Mitchell's $\subset$, since, for any object $A$, the identity $1_A$ is a monomorphism, so $A \subset A$.

*For the sake of my own sanity if nothing else, in this post I'll stick with Mitchell's notation and nomenclature as given above, even though I realize they may not be standard. Hence, inclusion, $\subset$, etc.

share|improve this question
1  
"Subobject" is supposed to be a generalization of "subobject" in the concrete sense in various concrete categories, including $\text{Set}$ but also including $\text{Grp}, \text{Top}, ...$. In these more complicated categories already the "problems" you describe exist. –  Qiaochu Yuan Sep 25 '11 at 0:34
1  
What I really have a problem with is this "inclusion" terminology, since it suppresses the choice of monomorphism and sounds like a binary relation, but isn't. –  Qiaochu Yuan Sep 25 '11 at 0:37
1  
It's known (see math.stackexchange.com/questions/55616/…) that $F_2$ contains a copy of $F_{\infty}$. So for 1) you can take $F_2$ and $F_{\infty}$, and for 2) you can take the identity map $F_2 \to F_2$ and the composite $F_2 \to F_{\infty} \to F_2$. These are "different" under any reasonable definition of different. –  Qiaochu Yuan Sep 25 '11 at 0:50
1  
But examples are easier to come up with in $\text{Top}$. For 1) consider $A = (0, 1), B = [0, 1]$ and for 2) consider for example the inclusions of a star with $3$ points into a graph containing at least two vertices of degree $3$ such that no automorphism of the graph permutes them. –  Qiaochu Yuan Sep 25 '11 at 0:53
1  
I believe the relevant ideas are all discussed in Munkres, esp. in Chapter 14. –  Qiaochu Yuan Sep 25 '11 at 18:17

3 Answers 3

up vote 3 down vote accepted

"Subobject" is supposed to generalize various concrete notions of "subobject," not only subset but also subgroup, subring, subspace, etc. Already for the case of subgroups and subspaces one encounters extra complications not found for subsets. For example:

  • It is possible for $A$ to be a subobject of $B$ and for $B$ to be a subobject of $A$ without it being the case that $A$ and $B$ are isomorphic. In other words, Cantor-Bernstein-Schroeder doesn't hold in general. For example $A = (0, 1), B = [0, 1]$ can be made subspaces of each other, but one is compact and one isn't. In $\text{Grp}$ we can take $A = F_2, B = F_{\infty}$ using the fact that $F_2$ contains $F_{\infty}$.
  • It is possible for $A$ to be a subobject of $B$ in "different ways." The strongest statement I could mean by this is that no automorphism of $B$ sends one monomorphism to another. In $\text{Grp}$ we can take the identity map $F_2 \to F_2$ and the composite $F_2 \to F_{\infty} \to F_2$; no automorphism exchanges these because the image of the second map has infinite index.

It's really these examples that you should be taking your intuition from. Also, you really shouldn't think of "subobject" as a relation: the crux of the matter is the monomorphism.

share|improve this answer
2  
A simple example of your second point is one-element subobjects in Top. There are plenty of topogical spaces whose automorphism group doesn't act transitively. –  Chris Eagle Sep 25 '11 at 20:06

"elements" are fundamental in set theory. However in category theory "maps" are fundamental. The concept of "subobject" is a category theoretic attempt at capturing the notion of subset, subgroup, subspace, etc. through maps.

It really is the best one can hope for. A category is a very general thing.

On the other hand if you know you have a concrete category (one built on sets), there might be a way to "fix" this notion of subobject. But then you really wouldn't be doing strict category theory anymore.

Somewhere buried in your question is the question, "Is category theory different from set theory?" The answer is "yes." If you look into Topoi theory you'll find that Topoi (special categories that "look" sort of like the category of sets) generally have a weaker version of internal logic than sets. So not as much can be said when working in this general context. http://en.wikipedia.org/wiki/Topos

share|improve this answer

This is one of those places where we blur the line between "is" and "is isomorphic to". For sets/groups/rings/etc., saying that there exists a monomorphism (i.e. an injection) $f:A\to B$ is equivalent to saying that $B$ contains and isomorphic copy of $A$, or that $B$ has a subobject isomorphic to $A$, or that $A$ can be embedded into $B$. The reason is that if you have an injective morphism $f:A\to B$, then the restricting the codomain to its image gives an isomorphism $\hat{f}:A\to \mathrm{Im}(f)$. Since two isomorphic objects are indistinguishable, if we know that $A$ can be embedded into $B$ then we can be lazy and regard it as a subobject of $B$*.

A shorter way to express this is that given a monomorphism, we identify the domain with the image which is both isomorphic to the domain and a true subobject of the codomain.

An example of this situation is the direct product of groups. Strictly speaking, under no circumstances is $A$ a subgroup of $A\times B$ (external direct product) and technically we should say $A\times \{id_B\}$ is the subgroup corresponding to $A$. The product can be described as the smallest group containing both $A$ and $B$, but this only makes sense if we regard $A$ as a subgroup.

Another example of this (in fact, one of my favorites) is saying $\mathbb{Q}$ is a subset of $\mathbb{R}$. When we define $\mathbb{R}$ axiomatically as the complete totally ordered field (link) there's no explicit reference to $\mathbb{Q}$ or how it fits in. As a field of characteristic zero, there's a unique way to embed $\mathbb{Q}$ into $\mathbb{R}$ that takes the unity of the first to the unity of the second. Once we note that we're free to talk about $\mathbb{Q}$ as sitting inside of $\mathbb{R}$ and write the real number corresponding to a/b as a/b as well.

When we generalize to categories, we lose the notions of injection and subobject (since the objects aren't necessarily sets anymore), but in many respects monomorphisms capture the useful properties of subobjects.

*We do have to be a little careful here if there's more than one way to embed $A$. (For example, given a group $G$ and two subgroups $H$ and $K$ which are isomorphic, then it's not necessarily true that $G/H$ is isomorphic to $G/K$). However, if we keep track of the monomorphism then there's no problem since we know exactly how $A$ fits in.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.