I am studying Awodey's Category theory book. I have trouble understanding the following line:
Observe that there are two functors in arrow category i.e.
$$
\begin{align}
\mathscr{C} \xleftarrow{\textbf{dom}} \mathscr{C}^{\rightarrow} \xrightarrow{\textbf{cod}} \mathscr{C}
\end{align}
$$
where $\mathscr{C}^{\rightarrow}$ is the arrow category corresponding to $\mathscr{C}$.
They have not mentioned what these $\textbf{dom}$ and $\textbf{cod}$ are? How do we prove that these are functors?
My understanding:
Now, in the diagram given below:
$\textbf{dom}:\mathscr{C}^{\rightarrow} \xrightarrow{\textbf{dom}} \mathscr{C}$. So,
$[f:A \to B] \mapsto A $ (object mapping of functor $\textbf{dom}$) and
if $g=(g_1,g_2):[f:A \to B] \to [f':A' \to B'] $, then $(g_1,g_2) \mapsto [f:A \to B]$ (the morphism mapping of functor $\textbf{dom}$ ).
Using this definition, If I proceed to prove the statement:
(a)
$$
\textbf{dom} ( g:f \to f' ) = \textbf{dom}(g): \textbf{dom}(f) \to \textbf{dom}(f')
$$
LHS = $f:A \to B$ and RHS = $\textbf{dom}(g): A \to A'$ (which seems absurd)
I am not sure if this makes sense.
Best Answer
Writing: $$f\stackrel{(g_1,g_2)}{\to}f'\tag1$$ where $f,f'$ are objects of arrow category $\mathcal C^{\to}$ and pair $(g_1,g_2)$ is an element of homset $\mathcal C^{\to}(f,f')$ represents a commuting diagram pictured in your question.
We have the functor $\mathbf{dom}:\mathcal C^{\to}\to\mathcal C$ prescribed by:$$[f\stackrel{(g_1,g_2)}{\to}f']\mapsto[\mathsf{dom}f\stackrel{g_1}{\to}\mathsf{dom}f']$$
And we have the functor $\mathbf{cod}:\mathcal C^{\to}\to\mathcal C$ prescribed by:$$[f\stackrel{(g_1,g_2)}{\to}f']\mapsto[\mathsf{cod}f\stackrel{g_2}{\to}\mathsf{cod}f']$$
In order to prove that $\mathbf{dom}$ and $\mathbf{cod}$ are functors it must be shown both of them respect identities and composition.
If $(1)$ stands for an identity then $f=f'$ and $g_1,g_2$ are both identities in $\mathcal C$. This guarantees that identities are respected.
By composition we must expand $(1)$ to: $$f\stackrel{(g_1,g_2)}{\to}f'\text{ and }f'\stackrel{(g'_1,g'_2)}{\to}f''\tag2$$with commuting squares.
Then we have: $$(g'_1,g'_2)\circ(g_1,g_2)=(g'_1\circ g_1,g'_2\circ g_2)$$assuring that composition is respected.