# How do morphisms in a comma category single out commuting squares?

I'm trying to teach myself the rudiments of Category Theory. I have a doubt about the definition of comma categories, more precisely about the morphisms.

Suppose have two functors $A\xrightarrow{S}C\xleftarrow{T}B$, and we construct the comma category $(S\downarrow T)$. A morphism between two objects $(\alpha,\beta,f)$ and $(\alpha',\beta',f')$ in the comma category is a pair $(g,h)$ where $g: \alpha \rightarrow \alpha'$ and $h : \beta \rightarrow \beta'$ are morphisms in $A$ and $B$ respectively, such that the following square commutes:

$\require{AMScd} \begin{CD} S(\alpha) @>{S(g)}>> S(\alpha')\\ @V{f}VV @VV{f'}V \\ T(\beta) @>{T(h)}>> T(\beta') \end{CD}$

My question is: how does $(g,h)$ single out a unique morphism $f'$ in $C$? Couldn't there be a second morphsim $f''$ which also made the square commute? But in that case $(g,h)$ wouldn't have a single target in $(S\downarrow T)$. Where am I going wrong?

• This question has been posed before and fully answered by David Moews and me: math.stackexchange.com/a/98358/19609 . Unfortunately the title of that question was very vague, so it is not obvious what it is all about, by just browsing the site index. Basically most/all textbooks use a sloppy notation for a morphism in a comma category. The correct one should be a quadruple, as described in the cited reference or in Berci's answer below. – magma Dec 29 '13 at 2:08

## 1 Answer

You're absolutely right in one sense: this phenomenon is analogous to that e.g. the function $x\mapsto x^2$ can be considered as an $\Bbb R\to\Bbb R$ function as well as an $\Bbb R\to\Bbb R^+$ function.

If you insist to separate the targets (and domains) for morphisms, then the proper definition of a morphism in the comma category would simply involve the target (and domain), so, if ${\bf f}=(\alpha,\beta,f)$ and ${\bf f'}=(\alpha',\beta',f')$ then a morphism of the comma category should be precisely defined as the 4-tuple $({\bf f},{\bf f'},g,h)$ with $g,h$ satisfying the commutative square condition.

However, this kind of things is not always handled so strictly, but it rarely would lead to any confusion, as when we write $(g,h):{\bf f}\to{\bf f'}$, the target and domain informations are already indicated...

• Yeah, it's the same way a set theoretic function, as a set of ordered pairs, has no unique codomain, but when you do category theory you assume a definition that does encode the codomain. Same with most morphisms in concrete categories. – Malice Vidrine Dec 28 '13 at 16:31