# Functors in arrow category

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.

• domain and codomain. Maps the arrow $f:A\to A'$ to $A$ and to $A'$ respectively. Commented Dec 31, 2018 at 8:26
• @LordSharktheUnknown I am not able to prove even, first condition of functor i.e. $\textbf{dom} ( g:f \to f' ) = \textbf{dom}(g): \textbf{dom}(f) \to \textbf{dom}(f')$
– MUH
Commented Dec 31, 2018 at 8:39

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.

The functor $$\mathbf{dom}$$ takes the object $$f$$ to $$A,$$ the object $$f'$$ to $$A';$$ and takes the arrow $$(g_1,g_2):f\to f'$$ to $$g_1.$$ Notice $$g_1$$ is an arrow $$A\to A'$$ 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 $$(g_1\circ h_1, g_2\circ h_2),$$ as can be seen by drawing two commuting squares side by side. So $$\mathbf{dom}((g_1,g_2)\circ (h_1,h_2)) = g_1\circ h_1 = \mathbf{dom}((g_1,g_2))\circ\mathbf{dom}((h_1\circ h_2))$$