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 $\mathbf C$; that is, $g \circ f$ is the unique element of $\mathbf C_1$ with $\map {R_\circ} {g, f, g \circ f}$.
== Proof ==
It 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:
:$\map {R_\circ} {x, \operatorname{dom} x, x}$
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 $\map {R_\circ} {x, \operatorname{dom} x, x}$ holds.
Axiom $(MOCT1)$ thus yields that:
:$\operatorname{cod} \operatorname{dom} x = \operatorname{dom} x$
as required.
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:
:$\map {R_\circ} {f, \operatorname{id}_{\operatorname{dom} f}, f}$
and:
:$\map {R_\circ} {\operatorname{id}_{\operatorname{cod} f}, f, f}$
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:
:$\map {R_\circ} {\operatorname{cod} f, f, f}$
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:
:$\map {R_\circ} {\operatorname{cod} f, f, f}$
which we have already seen to be an axiom.
The conditions:
:$\map {\operatorname{dom} } {g \circ f} = \operatorname{dom} f$ and $\map {\operatorname{cod} } {g \circ f} = \operatorname{cod} g$
follow literally from axiom $(MOCT2)$ and the definition of $g \circ f$.
The final associativity axiom:
:$h \circ \paren {g \circ f} = \paren {h \circ g} \circ f$
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
* Definition:Morphisms-Only Metacategory
* Characterization of Metacategory via Equations
Category:Category Theory