Dtyjlui

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.

Writing: $$fstackrel{(g_1,g_2)}{to}f'tag1$$ 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 $mathbf{dom}:mathcal C^{to}tomathcal C$ prescribed by:$$[fstackrel{(g_1,g_2)}{to}f']mapsto[mathsf{dom}fstackrel{g_1}{to}mathsf{dom}f']$$

And we have the functor $mathbf{cod}:mathcal C^{to}tomathcal C$ prescribed by:$$[fstackrel{(g_1,g_2)}{to}f']mapsto[mathsf{cod}fstackrel{g_2}{to}mathsf{cod}f']$$

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: $$fstackrel{(g_1,g_2)}{to}f'text{ and }f'stackrel{(g'_1,g'_2)}{to}f''tag2$$with commuting squares.

Then we have: $$(g'_1,g'_2)circ(g_1,g_2)=(g'_1circ g_1,g'_2circ g_2)$$assuring that composition is respected.

The functor $mathbf{dom}$ takes the object $f$ to $A,$ the object $f'$ to $A';$ and takes the arrow $(g_1,g_2):fto f'$ to $g_1.$ Notice $g_1$ is an arrow $Ato A'$ as it should be.

The identity morphism $mathrm{id}_f$ is simply $(mathrm{id}_A, mathrm{id}_{A'}),$ so $mathbf{dom}(mathrm{id}_f) = mathrm{id}_A$ as required.

The composition of two morphisms $(g_1,g_2)circ (h_1,h_2)$ is $(g_1circ h_1, g_2circ h_2),$ as can be seen by drawing two commuting squares side by side. So $mathbf{dom}((g_1,g_2)circ (h_1,h_2)) = g_1circ h_1 = mathbf{dom}((g_1,g_2))circmathbf{dom}((h_1circ h_2))$