Freyd: "is a subobject of" is not transitive

Question

On page 20 of Abelian Categories, Freyd writes

Note that the relation "is a subobject of" is not transitive.

On page 91 of Awodey's Category Theory (there are several typos in this page; the relevant one now is in the diagram - switch $M$ and $M^{\prime}$) we have that if $M^{\prime}\leq M$ and $M,M^{\prime}$ are subobjects of an object $X$, then the arrow $f$ in $m=m^{\prime}f$ (being monic since so are $m,m^{\prime}$) makes $M^{\prime}$ a subobject of $M$. Hence the subobject $M^{\prime}$ of a subobject $M$ of $X$ is itself a subobject of $X$.

Unless I am missing something, the only thing preventing a contradiction here is the premise $M^{\prime}\leq M$, which provides the monic $f$ along with the commuting triangle simultaneously makes $M^{\prime}$ a subobject of both $M$ (via $f$ itself), and $X$ (via $m^{\prime}$).

However, Awodey next writes we have a functor (there is another typo here, $M^{\prime}$ should be $M$ and the words "with $f$" should be erased) $$\mathsf{Sub}(M)\rightarrow \mathsf{Sub}(X)$$ Wouldn't such a functor make any subobject $M^{\prime}$ of $M$ a subobject of $X$, regardless of whether $M^{\prime}\leq M^{\prime}$ or not? Doesn't this contradict Freyd's statement?

Answer 1

(In my version of Awodey this is page 78-9, not 91.)

Freyd is just making a linguistic point, as made clear by what he says next: "Indeed, subobjects, as we have defined them, do not have subobjects" -- only objects have subobjects. A subobject is not an object, but rather a monomorphism (or, depending on your definition, an equivalence class of monomorphisms). So you can't even state transitivity, strictly speaking. As Freyd puts it, "This is a baroque consideration".

The transitivity result that I think you're driving at is that if $M' \overset{f}{\to}M$ is a subobject of $M$ and $M \overset{m}{\to} X$ is a subobject of $X$, then since monomorphisms are closed under composition, $M' \overset{m \circ f}{\to} X$ is a subobject of $X$. By abuse of language, omitting mention of the arrows involved, one might just say: if $M'$ is a subobject of $M$ and $M$ is a subobject of $X$, then $M'$ is a subobject of $X$. But technically it's only an abuse of language that allows one to say this.

Answer 2

In my opinion, this is just Freyd's cheeky way of saying that "subobject" is overloaded. It both means:

1. An object of the category $\mathrm{Sub}_X$ whose objects are monomorphisms into $X$, and whose arrows are commutative triangles. (Exercise; show that there is at most one arrow between any two objects of this category.)
2. A binary relation $\subseteq_X\, : \mathrm{Sub}_X \times \mathrm{Sub}_X \rightarrow \{False,True\},$ expressing that one arrow factorizes through the other (which, on the basis of the previous exercise, completely determines the structure of $\mathrm{Sub}_X$.)

These are related, of course. Given an object $Y$, there is an "inclusion" $\eta_{X,Y} : \mathrm{Mono}(Y,X) \rightarrow \mathrm{Sub}_X$. (I use scare-quotes because $\eta_{X,Y}$ needn't be injective; by which I mean simply that we can have distinct monomorphisms $Y \rightarrow X$ mapped to isomorphic objects of $\mathrm{Sub}_X.$) If we have arrows $f : A \rightarrow X$ and $g : B \rightarrow X$ whose images under $\eta$ satisfy $$\eta_{X,A}(f) \subseteq_X \eta_{X,B}(g),$$

then there is a unique arrow $c : A \rightarrow B$ such that $g \circ c = f$. Now it is a general principle that if a composite $g \circ c$ is a monomorphism, then $c$ is also a monomorphism (exercise.) It follows that $\eta_{B,A}(c)$ is well-defined, so we've got ourselves a subobject of $B$ that could reasonably be called "$A$ viewed as a subobject of $B$."