TITLE: Question about general comma categories
QUESTION [6 upvotes]: 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, 
$$(A,f,B)\xrightarrow{(\phi,\psi)}(C,g,D)\quad\text{and}\quad(A,f',B)\xrightarrow{(\phi,\psi)}(C,g',D).$$

REPLY [5 votes]: 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