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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?

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In the correct version of set theory (say, ETCS), it isn't even possible to ask this question. "Disjoint" is a property of subsets of a set, not a property of two sets. –  Qiaochu Yuan Mar 31 at 5:46
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@QiaochuYuan There are some respectable versions of the definition of category that begin with a set (or object) of objects, a set (or object) of morphisms, maps assigning to each morphism a source and a target, etc. For example, this seems to be the standard way to define internal categories of a topos. And in such a context, it is meaningful and true to say that the Hom-sets are disjoint. –  Andreas Blass Mar 31 at 14:25
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3 Answers 3

up vote 1 down vote accepted

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.

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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 $(\operatorname{id}_x,\operatorname{id}_y)\colon g\to g$.

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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.

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