Characterization of Metacategory via Equations

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Theorem

Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.

Let $\operatorname{Cdm}$ and $\operatorname{Dom}$ assign to every element of $\mathbf C_1$ an element of $\mathbf C_0$.

Let $\operatorname{id}$ assign to every element of $\mathbf C_0$ an element of $\mathbf C_1$.

Denote with $\mathbf C_2$ the collection of pairs $\tuple {f, g}$ of elements of $\mathbf C_1$ satisfying:

$\Dom g = \Cdm f$

Let $\circ$ assign to every such pair an element of $\mathbf C_1$.

Then $\mathbf C_0, \mathbf C_1, \operatorname{Cdm}, \operatorname{Dom}, \operatorname{id}$ and $\circ$ together determine a metacategory $\mathbf C$ if and only if the following seven axioms are satisfied:

where $A$ and $f, g, h$ are arbitrary elements of $\mathbf C_0$ and $\mathbf C_1$, respectively.

Further, in the last two lines, it is presumed that all compositions are defined.

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Hence it follows that:

• $\mathbf C_0$ and $\mathbf C_1$ represent the collections of objects and morphisms of $\mathbf C$
• $\operatorname{Dom}$ and $\operatorname{Cdm}$ represent the domain and codomain of a morphism of $\mathbf C$
• $\operatorname{id}$ represents the identity morphisms of $\mathbf C$
• $\mathbf C_2$ represents the collection of composable morphisms of $\mathbf C$
• $\circ$ represents the composition of morphisms in $\mathbf C$

Proof

Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.

Let $A$ in $C_O$ be arbitrary. Let $f, g, h$ in $C_1$ be arbitrary.

By definition of identity morphism:

$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$

$f \circ \operatorname{id}_A = f$

$\operatorname{id}_A \circ g = g$

whenever $A$ is the domain of $f$ or the codomain of $g$, respectively.

From definition of identity morphism, since:

$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$:

Therefore:

$\Dom{\operatorname{id}_A} = A$

$\Box$

Similarly, from definition of identity morphism, since:

$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$:

Therefore:

$\Cdm{\operatorname{id}_A} = A$

$\Box$

From definition of identity morphism again, since:

$f \circ \operatorname{id}_A = f$

and:

$\Dom{\operatorname{id}_A} = A$

Therefore:

$f \circ \operatorname{id}_A = f \circ \operatorname{id}_{\Dom f} $

Hence:

$f \circ \operatorname{id}_{\Dom f} = f$

$\Box$

From definition of identity morphism again, since:

$\operatorname{id}_A \circ g = g$

and:

$\Cdm{\operatorname{id}_A} = A$

Therefore:

$\operatorname{id}_A \circ f = \operatorname{id}_{\Cdm f} \circ f $

Hence:

$\operatorname{id}_{\Cdm f} \circ f = f$

$\Box$

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Sources

• 2010: Steve Awodey: Category Theory (2nd ed.) ... (next): $\S 3.1$