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Let $F:\cal{A}\to\cal{C}$ and $G:\cal{B}\to\cal{C}$ be functors, and let $(F\downarrow G)$ be the comma category of $F$ and $G$.

My question is, how do we know that the Hom-sets are pariwise disjoint? The only case I am struggling with is where we have the four objects: $$(A,f,B)\quad (C,g,D)$$ $$(A,f',B)\quad (C,g',D)$$ with $f\neq f'$ and $g\neq g'$. To be more explicit, how do we know that there are no pairs of morphisms $(\phi,\psi)$ such that both squares $$ \require{AMScd} \begin{CD} FA @>{F\phi}>> FC\\ @V{f}VV @VV{g}V \\ GB @>{G\psi}>> GD \end{CD}\quad\quad\quad \begin{CD} FA @>{F\phi}>> FC\\ @V{f'}VV @VV{g'}V \\ GB @>{G\psi}>> GD \end{CD} $$ commute? That is, $$(A,f,B)\xrightarrow{(\phi,\psi)}(C,g,D)\quad\text{and}\quad(A,f',B)\xrightarrow{(\phi,\psi)}(C,g',D).$$

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    I don't have an example, but I suspect that such pairs do exist in general. Why do you want them not to exist? – Jim Jan 26 '15 at 04:05
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    For a simple example of such a pair choose everything such that $F\phi$ and $G\psi$ are isomorphisms. Then for any choice of $f$ and $f'$ there exist choices for $g$ and $g'$ that make the diagrams commute. – Jim Jan 26 '15 at 04:10
  • @Jim That would imply that the Hom sets were not pairwise disjoint, which is required in order to have a category (at least in my book). –  Jan 26 '15 at 04:12
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    Generally homs are considered to be adorned by their domain and codomain, so different homsets for different objects are just defined to be different sets. – Jim Jan 26 '15 at 04:14
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    For example I can define a category by saying the objects are a, b, c and the morphisms are elements of $\mathbb Z$ and composition is multiplication. Even though every hom sets has a $2$ we still think of $2\colon a \to b$ and $2\colon b \to c$ to be distinct elements because their domain and ranges are not equal. – Jim Jan 26 '15 at 04:15
  • @Jim OH! Wow that never occurred to me. That clears up everything. Thanks for the help! –  Jan 26 '15 at 04:20
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    Just to emphasize Jim's comment : would you say that the free product of groups $\mathbb Z/2 \ast \mathbb Z/2$ is not well-defined because each component got an element called $1$ ? Of course not, you would say "let's denote ${0,1_l}$ the left copy and ${0,1_r}$ the right copy and then a word in the free group admits a unique reading". This is exactly the same here : if you really want to make it formal, you can say that a morphism in a category is a triple $(a,b,f)$ where $a$ and $b$ are objects. – Pece Jan 26 '15 at 09:05
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    This is relevant. In general it's not very important (or important at all) to have disjoint hom-sets. – Najib Idrissi Jan 26 '15 at 14:36

1 Answers1

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Welcome to the club! the club of (relatively) few people who realized that the standard textbook notation of a comma category morphism as an ordered pair ($(\phi,\psi)$ in your example) is incomplete/incorrect.

To make a short story even shorter: the notation for comma category morphisms should be a quadruple, not a couple. For example - in your left square diagram - the morphism (if it exists) should be written $(f,\phi,\psi,g)$, while the one on the right square diagram (if it exists) should be written $(f',\phi,\psi,g')$, where the first and last entries in the quadruples indicate the domain and codomain (in the comma category) of the comma category morphism (obviously a different order in the quadruple items is possible, as long as you explain it to your readers ).

You can read more about this (and meet the other members of the club) by reading the post and my answer here

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magma
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  • Ha, thanks! I think I see now. So, if I require that the Hom sets be disjoint I need to define this quadruple in order to avoid these issues. However, if I define morphisms (in an arbitrary category) to be triples $(A,f,B)$ as Jim and Pece said above, I'll be alright? –  Jan 26 '15 at 15:24
  • Look, the basic rule is: different things should be written in different ways, so you can see that they are different (and you can make calculations, manually or automatically with a PC). You can of course write morphisms as triples, but that is overkill in 99% of the cases. You would than need a quadruple of triples to describe a morphism in a comma category. So you do that only in special cases as in @Jim example with Z, where you cannot freely choose the names of the morphisms. continued below – magma Jan 26 '15 at 17:09
  • continued from above. Comma cats are one such case: you already have named morphisms in the base categories, so you need a precise notation to allow you to recover the domain and codomain information. That's why you need a quadruple. – magma Jan 26 '15 at 17:10
  • Gotcha. Thanks for the help! –  Jan 26 '15 at 17:17
  • you are welcome and if you found my other answer valuable you kindly are invited to upvote it. – magma Jan 26 '15 at 17:55
  • please see my answer below, I may have found a possible issue with the quadruple definition. –  Jan 27 '15 at 23:38
  • Wow. Can't believe I missed that. Thanks again. Also, just out of curiosity, how do you feel about Jim's and Najib's comment above? I'm getting mixed statements about how important this disjointness condition is. –  Jan 28 '15 at 16:43
  • @CC0607 the disjointness condition is VERY important, but it is enforced stating that a category comes with cod and dom functions associating to each morphism a domain and a codomain. Najib points you to a relevant q&A but then - inexplicably to me - comes to the opposite conclusion. Remember his reference, forget his comment. Cont.d – magma Jan 28 '15 at 20:08
  • Jim's a,b,c Z construction is - very strictly speaking - not a category because the dom and cod functions are undefined. But of course anybody with a minimum of mathematical experience (or simple common sense) can figure out that all you need to upgrade the structure to the category status is just to write $<a,n,b>$ and $<b,n,c>$ to distinguish different morphisms which were initially just called $n$. Cont.d – magma Jan 28 '15 at 20:09
  • I suggest you to use the Wikipedia Definition of category which does not talk about hom disjointness, but it derives it as a consequence. – magma Jan 28 '15 at 20:11
  • Gotcha. Thanks for being patient with me! –  Jan 28 '15 at 20:27