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Stripping away the bulk of category theory and treating [small] categories as a type of algebraic structure, does the following suffice to axiomatise category theory? If not, what am I missing?

A [small] category is a $5$-tuple $\mathcal{C}=(C,M,s,t,\circ)$, where $C$ is the set of objects, $M$ the set of morphisms, $s:M\to C$ the source function, $t: M\to C$ the target function, and $\circ$ a partial operation $M^2\to M$, which satisfies the category axioms$^*$:

$$\begin{matrix} \mathbf{comp}_1 & \forall f,g\in M.t(f)=s(g)\implies g\circ f\in M\\ \mathbf{comp}_2 & \forall f,g\in M.g\circ f\in M\implies s(g\circ f)=s(f)\\ \mathbf{comp}_3 & \forall f,g\in M.g\circ f\in M\implies t(g\circ f)=t(g)\\ \mathbf{assoc} & \forall f,g,h\in M.f\circ(g\circ h)=(f\circ g)\circ h\\ \mathbf{id} & \forall X\in C.\exists f\in M.s(f)=t(f)=X \end{matrix}$$

Edit:

$$\begin{matrix} \mathbf{id}_1 & \forall X\in C.\exists f\in M.s(f)=t(f)=X\\ \mathbf{id}_2 & \forall f,g\in M.s(f)=t(f)\land s(g)=t(f)\implies g\circ f=g\\ \mathbf{id}_3 & \forall f,g\in M.s(f)=t(f)\land t(g)=s(f)\implies f\circ g=g \end{matrix}$$

$^*$ In keeping with the typical naming conventions of abstract algebra (e.g. group axioms, field axioms, etc.)

R. Burton
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    The composition is only a partial operation, not (necessarily) defined on whole $M^2$. What is your goal? – Berci Mar 03 '20 at 18:03
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    The composition should only be partially defined (in the style of your axiom 1), but you declare composition as a function on all pairs of morphisms. – Captain Lama Mar 03 '20 at 18:03
  • @Berci the goal is 1) to describe category theory in terms of abstract algebra and 2) to identify structures which are categories (in that they satisfy the "category axioms"). The second part is because I noticed that certain structures which wouldn't ordinarily be considered "categories," can be interpreted as categories via the above. – R. Burton Mar 03 '20 at 18:10
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    Your id postulate is not enough - you need that the $f$ acts as and id: $f\circ g=g$ and $g\circ f=g$ when these are defined. – Thomas Andrews Mar 03 '20 at 18:12

1 Answers1

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After noting that objects are unnecessary (we can just identify them with the identity arrows)$^1$, there are three basic ways to treat (small) categories "algebraically:"

  • Essentially as you've done (after throwing out objects and more importantly incorporating Thomas Andrews' observation in the comments). This involves using partial functions - namely composition - which isn't always something we want to do; on the plus side, it's extremely natural.

  • Using relations to replace the partial composition function. This is the approach I've seen in the literature, and I mention it here. The downside is that this is no longer strictly "algebraic" in the sense of universal algebra.

  • Adding some "formal element" $\perp$ which we send every undefined expression to and which is an annihilator for every operation (e.g. "composing" $\perp$ with something just yields $\perp$ again). This carries the same information but is generally considered bad, if only from an aesthetic standpoint.

$^1$See my linked answer for how this plays out if we use the second approach; it's handled identically in the first and third.

Noah Schweber
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  • This gives me an interesting idea. What if you identify each morphism with a point in $X^3$ (for some appropriate set $X$), and treat the homset as a projection onto $X^2$? Could you use geometry to infer information about the category and vice-versa? – R. Burton Mar 03 '20 at 18:37
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    @R.Burton I don't see what $X$ should be here. If the homset is given by projection to $X^2$ then presumably $X$ is the set of objects (with $proj^{-1}(\langle a,b\rangle)=Hom(a,b)$), but consider e.g. a category with one element and infinitely many morphisms. We do have a map from the set of all arrows to the set of ordered pairs of objects, but that's quite different from what you've written. – Noah Schweber Mar 03 '20 at 18:40
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    Regardless, you're not going to get any nontrivial information out of this sort of thing - once appropriately patched, it will be just a direct rephrasing of what's already given. – Noah Schweber Mar 03 '20 at 18:40
  • In this case each arrow is triple, with the third (or first coordinate) being used to distinguish between morphisms and the remaining two being the source and target. $X$ is indeed the set of objects. In a category with one element and infinitely many morphisms, every morphism would be projected to $(x,x)$ (where $x$ is the sole object of $X$). This isn't particularly interesting, though. – R. Burton Mar 03 '20 at 18:49
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    @R.Burton My point is that there aren't enough third coordinates to account for all the arrows. E.g. in a one-element category with element $x$, every arrow would have to be represented by the unique triple $(x,x,x)$; but as soon as there are two distinct arrows - let alone infinitely many - we can't do that (unless we're allowing the same triple to represent multiple arrows, in which case ... why are we doing this?). – Noah Schweber Mar 03 '20 at 18:50
  • Mostly for the fun of it. But I guess you're right, it should be $X^2\times Y$. – R. Burton Mar 03 '20 at 19:19