Showing hom-sets are disjoint in a morphism category

Question

Let $\mathcal{C}$ be a category. We can construct a category of morphisms $\mathcal{M}$ by letting the objects be morphisms in $\mathcal{C}$ and morphisms be appropriate pairs of morphisms that give a commutative diagram.

Let $f:A \to B$, $g:C \to D$, $h:E \to F$, and $k:G \to H$. A morphism $(r,s) \in \mathrm{Hom}(f,g)$ consists of a map $r:A \to C$ and a map $s:B \to D$, and a morphism $(t,u) \in \mathrm{Hom}(h,k)$ consists of a map $t:E \to G$ and a map $u:F \to H$, all so that the appropriate diagrams commute.

Part of the requirement of a category is that hom-sets be disjoint. Suppose $(r,s)=(t,u)$ for some $(t,u) \in \mathrm{Hom}(h,k)$. Then we must have $r:E \to G$ and $s:F \to H$, so it must be that $A=E$, $C=G$, $B=F$, and $D=H$.

Since the morphisms $r:A \to C$ and $s:B \to D$ form a commutative diagram with $f:A \to B$ and $g:C \to D$, we have the relations $k \circ r=s \circ h$ and $g \circ r=s \circ f$. I'm not sure how to show from here that $r=t$ and $s=u$, which would show that if two hom-sets are not equal then they are disjoint. I see that they must have the same domain and codomain, but I feel like this isn't enough.

How do we show that hom-sets in a morphism category are disjoint?

Answer 1

You're problem is due to the fact that the correct definition of the arrow-category (of a given category $\mathbf C$) requires that a morphism from $f \colon X \to Y$ and $g \colon Z \to W$, let's call it $\sigma \colon f \to g$, should be a commutative square: a $4$-tuple $\sigma = \langle f,g,k,h\rangle$ with $k \colon X \to Z$ and $h \colon Y \to W$ such that the induced diagrams commute.

The point is that arrow of a category should carry information about its domain and its codomain, and with the definition above every arrow have a unique domain e codomain: it could be possible that there are two $4$-ple $\langle f,g,k,h\rangle$ and $\langle a,b,k,h\rangle$ which are arrows in a category, anyway these are different arrows because they have different source and target (namely $f$ and $g$ in the first case and $a$ and $b$ in the second one).

This is the same problem you have in $\mathbf {Set}$ if you reguard morphism (i.e. functions) as just relations (i.e. subsets of cartesian products): if you consider $f \colon X \to Y$ to simply be a functional relation $f \subseteq X \times Y$ then, if $f$ isn't sujective, it cannot be distingueshed from any other function $f' \subseteq X \times Y'$ obtained restricting the codomain.

Mac Lane solved this problem in the introduction of Categories for working Mathematicians by defining a function $f \colon X \to Y$ to be a triple $\langle X, Y, F\rangle$ where $F \subseteq X \times Y$ is a functional relation.

Hope this helps.

Answer 2

As stated in the comment by Qiaochu Yuan, the homsets of a category should be considered disjoint by definition. Since for any pair of objects $x$, $y$ in a category $\mathcal{C}$, you define a set $\operatorname{Hom}(x,y)$. If you want to define $\operatorname{Hom}(x,y)$ as a subset of $\operatorname{Mor}\mathcal{C}$, you need to provide additional data, i.e. domain and codomainmaps $\operatorname{dom},\operatorname{cod}\colon\operatorname{Mor}\mathcal{C}\to\operatorname{Ob}\mathcal{C}$

Beside that, your claim that the homsets of the morphismcategory $\mathcal{M}$ are disjoint subsets of $\operatorname{Mor}\mathcal{C}\times \operatorname{Mor}\mathcal{C}$ is wrong. Consider two parallel morphisms $f,g\colon x\to y$. Then $(\operatorname{id}_x,\operatorname{id}_y)$ would be a morphism in $\operatorname{Hom}(f,f)$, and a morphism $\operatorname{Hom}(g,g)$. To distinguish them, you need to provide additional data, i.e. the domain and codomain of the respective morphisms (or equivalently, the knowledge to which homset they belong), such that $(\operatorname{id}_x,\operatorname{id}_y)\colon f\to f$ is a different morphism than $(\operatorname{id}_x,\operatorname{id}_y)\colon g\to g$.

Answer 3

Since the "morphism category" (the more usual name is "arrow category" though) of $C$ is nothing but a special type of comma category - the category $\left(\text{Id}_C\downarrow \text{Id}_C\right)$ - this question has been exaustively answered a long time ago by myself and others here and more recently here.

In short: what you read in most/all category books that: a morphism in the arrow category or in a comma category is a pair of arrows....is simply not correct. You should have a quadruple of arrows (2 arrows being the domain and the codomain).

Let's just hope more modern category theory books will be more precise.

Answer 4

Homsets aren't required to be disjoint: insisting on it is just a technical convenience for working with a notion of sets that allows you to ask such things of unrelated sets. (such as ZFC and its global membership relation)

For example, one can (and often does; e.g. to talk about a category enriched in Set) define a small category to be a set of objects $O$, a set-valued $\hom$ on $O \times O$, and a set of functions $\hom(b,c) \times \hom(a,b) \to \hom(a,c)$ satisfying properties.

It's easy to see that, with the notion of category so defined, any nonempty small category is equivalent to one with disjoint homsets, and also equivalent to one whose homsets are not disjoint.

Requiring that homsets be disjoint lets us do things like take their union to get the set of all arrows; otherwise to do such things we'd have to work with a disjoint union, and have to deal with all of the inconveniences that entails.

There isn't really anything to be gained by fixating on this detail though; it's sort of like asking whether $3$ is an element of $\pi$; while it is well-defined to ask such a question about an interpretation of the real numbers in sets, you're probably not doing things right if the answer to the question actually matters to you.