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:

enter image description here

$\textbf{dom}:\mathscr{C}^{\rightarrow} \xrightarrow{\textbf{dom}} \mathscr{C}$. 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.

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

2 Answers 2


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 $\mathbf{dom}((g_1,g_2)\circ (h_1,h_2)) = g_1\circ h_1 = \mathbf{dom}((g_1,g_2))\circ\mathbf{dom}((h_1\circ h_2))$


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