Morphisms-Only Metacategory Induces Metacategory

Theorem
Let $\mathbf C$ be a morphisms-only metacategory.

Then $\mathbf C$ induces a metacategory $\mathbf C'$, as follows (phrased to fit with Characterization of Metacategory via Equations):


 * Define $\mathbf C'_1$ to be the collection $\mathbf C_1$ of morphisms of $\mathbf C$.


 * Define $\mathbf C'_0$ to be the image of the operation $\operatorname{dom}$ on $\mathbf C_1$.


 * Define $\operatorname{id}$ to be the identity on $\mathbf C'_0$, and take $\operatorname{dom}$ and $\operatorname{cod}$ as in $\mathbf C$.


 * Define $\circ$ to be as in $\mathbf C$; i.e., $g \circ f$ is the unique element of $\mathbf C_1$ with $R_\circ \left({g, f, g \circ f}\right)$.

Proof
It just remains to verify that these definitions satisfy the premises for applying Characterization of Metacategory via Equations.

The first of these is the assertion:


 * $\operatorname{dom} \operatorname{id}_A = A$

Here, $A \in \mathbf C'_0$, so by above definition there is some $x \in \mathbf C_1$ with $A = \operatorname{dom} x$.

Then also $\operatorname{id}_A = \operatorname{dom} x$; hence it suffices to verify:


 * $\operatorname{dom} \operatorname{dom} x = \operatorname{dom} x$

From axiom $(MOCT4)$, we have $R_\circ \left({x, \operatorname{dom} x, x}\right)$.

From axiom $(MOCT2)$, this implies $\operatorname{dom} x = \operatorname{dom} \operatorname{dom} x$.

Now, to verify:


 * $\operatorname{cod} \operatorname{id}_A = A$

Reasoning as above, this comes down to proving:


 * $\operatorname{cod} \operatorname{dom} x = \operatorname{dom} x$

Axiom $(MOCT4)$ gives again that $R_\circ \left({x, \operatorname{dom} x, x}\right)$ holds.

Axiom $(MOCT1)$ thus yields that $\operatorname{cod} \operatorname{dom} x = \operatorname{dom} x$, as desired.

For the next two conditions, we have to verify:


 * $f \circ \operatorname{id}_{\operatorname{dom} f} = f$ and $\operatorname{id}_{\operatorname{cod} f} \circ f = f$

Thus, rephrasing in the language of the morphisms-only metacategory $\mathbf C$, this means:


 * $R_\circ \left({f, \operatorname{id}_{\operatorname{dom} f}, f}\right)$ and $R_\circ \left({\operatorname{id}_{\operatorname{cod} f}, f, f}\right)$

By definition of $\operatorname{id}$, the first of these is contained in axiom $(MOCT4)$.

For the second one, it is first to be shown that $\operatorname{cod} f$ is a legitimate object of $\mathbf C'_0$.

Axiom $(MOCT4)$ again yields us that $R_\circ \left({\operatorname{cod} f, f, f}\right)$, from which by axiom $(MOCT2)$ it follows that:


 * $\operatorname{cod} f = \operatorname{dom} \operatorname{cod} f$

Now by definition of $\operatorname{id}$ the second statement can be reduced to $R_\circ \left({\operatorname{cod} f, f, f}\right)$ which we already saw to be an axiom.

The conditions:


 * $\operatorname{dom} \left({g \circ f}\right) = \operatorname{dom} f$ and $\operatorname{cod} \left({g \circ f}\right) = \operatorname{cod} g$

follow literally from axiom $(MOCT2)$ and the definition of $g \circ f$.

The final associativity axiom:


 * $h \circ \left({g \circ f}\right) = \left({h \circ g}\right) \circ f$

is phrased by $(MOCT3)$.

Having verified all seven conditions, we conclude that $\mathbf C'$ with the defined operations constitutes a metacategory.

Also see

 * Metacategory Induces Morphisms-Only Metacategory
 * Morphisms-Only Metacategory
 * Characterization of Metacategory via Equations