functor precategory (Rev #4, changes)

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For precategories $A,B$, there is a precategory $B^A$, called the **functor precategory**, defined by

- $(B^A)_0$ is the type of functors from $A$ to $B$.
- $hom_{B^A}(F,G)$ is the type of natural transformations from $F$ to $G$.

**Proof.** We define $(1_F)_a \equiv 1_{F a}$. Naturality follows by the unit axioms of a precategory. For $\gamma : F \to G$ and $\delta : G \to H$, we define $(\delta \circ \gamma)_a \equiv \delta_a \circ \gamma_a$. Naturality follows by associativity. Similarly, the unit and associativity laws for $B^A$ follow from those for $B$.

A natural transformation $\gamma : F \to G$ is an isomorphism in $B^A$ if and only if each $\gamma_a$ is an isomorphism in $B$.

**Proof.** If $\gamma$ is an isomorphism?, then we have $\delta : G \to F$ that is its inverse. By definition of composition in $B^A$, $(\delta \gamma)_a \equiv \delta_a \gamma_a$. Thus, $\delta \gamma = 1_F$ and $\gamma \delta=1_G$ imply that $\delta_a \gamma_a = 1_{F a}$ and $\gamma_a \delta_a = 1_{G a}$, so $\gamma_a$ is an isomorphism.

Conversely, suppose each $\gamma_a$ is an isomorphism?, with inverse called $\delta_a$. We define a natural transformation $\delta : G \to F$ with components $\delta_a$; for the naturality axiom we have

$F f \circ \delta_a=\delta_b \circ \gamma_b \circ F f \circ \delta_a = \delta_b \circ G f \circ \gamma_a \circ \delta_a = \delta_b \circ G f$

Now since composition and identity of natural transformations is determined on their components, we have $\gamma \delta=1_G$ and $\delta \gamma 1_F.\ \square$

If $A$ is a precategory and $B$ is a category, then $B^A$ is a category.

**Proof.** Let $F,G:A\to B$; we must show that $idtoiso\mathrm{idtoiso}:(idF=G)\to (F\cong G)$~~ \idtoiso:(\id{F}{G})~~ idtoiso:({F}={G}) \to (F\cong G) is an~~ equivalence.~~equivalence.

To give an inverse to it, suppose $\gamma:F\cong G$ is a~~ natural~~~~ isomorphism.~~~~ Then~~~~ for~~~~ any~~~~$a:A$~~natural isomorphism~~ ,~~ .~~ we~~ Then~~ have~~ for~~ an~~ any~~ isomorphism~~${\gamma}_{a}a:\mathrm{FaA}\cong \mathrm{Ga}$~~ \gamma_a:Fa~~ a:A~~ \cong~~~~ Ga~~ ,~~ hence~~ we have an~~ identity~~~~$\isotoid(\gamma_a):\id{Fa}{Ga}$~~isomorphism?~~ .~~~~ By~~~~ function~~~~ extensionality,~~~~ we~~~~ have~~~~ an~~~~ identity~~$\overline{\gamma}{\gamma}_{a}:idF[a(\cong {A}_{0}G\to a{B}_{0})]{F}_{0}{G}_{0}$~~ \bar{\gamma}:\id[(A_0\to~~ \gamma_a:F~~ B_0)]{F_0}{G_0}~~ a \cong G a, hence an identity $isotoid(\gamma_a):{F a}={G a}$. By function extensionality, we have an identity $\bar{\gamma}:{F_0}=_{(A_0\to B_0)}{G_0}$.

Now since the last two axioms of a~~ functor~~~~ are~~~~ mere~~~~ propositions,~~~~ to~~~~ show~~~~ that~~~~$\id{F}{G}$~~functor ~~ it~~ are~~ will~~~~ suffice~~~~ to~~~~ show~~~~ that~~~~ for~~~~ any~~~~$a,b:A$~~mere propositions?, to show that ${F}={G}$ it will suffice to show that for any $a,b:A$, the functions

```
F_{a,b} hom_A(a,b) \to hom_B(Fa,Fb)\mathrlap{\qquad\text{and}}\\
G_{a,b}&:hom_A(a,b) \to hom_B(Ga,Gb)
```

$G_{a,b}:hom_A(a,b) \to hom_B(G a,G b)$

become equal when transported along $F_{a,b} hom_A(a,b) \to hom_B(Fa,Fb)\mathrlap{\qquad\text{and}}\\\bar\gamma$. By computation for function extensionality, when applied to $a$, $\bar\gamma$ becomes equal to $\isotoid(\gamma_a)$. But by \cref{ct:idtoiso-trans}, transporting $Ff:hom_B(Fa,Fb)$ along $\isotoid(\gamma_a)$ and $\isotoid(\gamma_b)$ is equal to the composite $\gamma_b\circ Ff\circ \inv{(\gamma_a)}$, which by naturality of $\gamma$ is equal to $Gf$.

become equal when transported along $\bar\gamma$. By computation for function extensionality, when applied to $a$, $\bar\gamma$ becomes equal to $isotoid(\gamma_a)$. But by \cref{ct:idtoiso-trans}, transporting $F f:hom_B(F a,F b)$ along $isotoid(\gamma_a)$ and $isotoid(\gamma_b)$ is equal to the composite $\gamma_b\circ F f\circ \inv{(\gamma_a)}$, which by naturality of $\gamma$ is equal to $G f$.

This completes the definition of a function $(F\cong G) \to (\id F G)$. Now consider the composite

This completes the definition of a function $(F\cong G) \to (F =G)$. Now consider the composite

(1)$(\id F G) \to (F\cong G) \to (\id F G).$

$(F= G) \to (F\cong G) \to (F =G).$

Since hom-sets are sets, their identity types are mere propositions, so to show that two identities $p,q:\id F G$ are equal, it suffices to show that $\id[\id{F_0}{G_0}]{p}{q}$. But in the definition of $\bar\gamma$, if $\gamma$ were of the form $\idtoiso(p)$, then $\gamma_a$ would be equal to $\idtoiso(p_a)$ (this can easily be proved by induction on $p$). Thus, $\isotoid(\gamma_a)$ would be equal to $p_a$, and so by function extensionality we would have $\id{\bar\gamma}{p}$, which is what we need.

Since hom-sets are sets, their identity types are mere propositions, so to show that two identities $p,q:F =G$ are equal, it suffices to show that $p =_{{F_0}={G_0}}{q}$. But in the definition of $\bar\gamma$, if $\gamma$ were of the form $idtoiso(p)$, then $\gamma_a$ would be equal to $idtoiso(p_a)$ (this can easily be proved by induction on $p$). Thus, $isotoid(\gamma_a)$ would be equal to $p_a$, and so by function extensionality we would have ${\bar\gamma}={p}$, which is what we need.

Finally, consider the composite

Since identity of~~ natural~~~~ transformations~~~~ can~~~~ be~~~~ tested~~~~ componentwise,~~~~ it~~~~ suffices~~~~ to~~~~ show~~~~ that~~~~ for~~~~ each~~~~$a$~~natural transformations ~~ we~~ can~~ have~~ be tested componentwise, it suffices to show that for each$idaidtoiso(\overline{\gamma}{)}_{a}{\gamma}_{a}$~~ \id{\idtoiso(\bar\gamma)_a}{\gamma_a}~~ a~~ .~~ ~~ But~~~~ as~~~~ observed~~~~ above,~~ we have$id\mathrm{idtoiso}(\overline{\gamma}{)}_{a}idtoiso(\overline{\gamma}{)}_{a}=idtoiso{\gamma}_{a}((\overline{\gamma}{)}_{a})$~~ \id{\idtoiso(\bar\gamma)_a}{\idtoiso((\bar\gamma)_a)}~~ {idtoiso(\bar\gamma)_a}={\gamma_a}~~ ,~~ .~~ while~~ But as observed above, we have$id\mathrm{idtoiso}(\overline{\gamma}{)}_{a}(\overline{\gamma}{)}_{a}=isotoid\mathrm{idtoiso}({\gamma}_{a}(\overline{\gamma}{)}_{a})$~~ \id{(\bar\gamma)_a}{\isotoid(\gamma_a)}~~ {idtoiso(\bar\gamma)_a}={idtoiso((\bar\gamma)_a)}~~ ~~ ,~~ by~~ while~~ computation~~~~ for~~~~ function~~~~ extensionality.~~~~ Since~~$isotoid(\overline{\gamma}{)}_{a}=\mathrm{isotoid}({\gamma}_{a})$~~ \isotoid~~ {(\bar\gamma)_a}={isotoid(\gamma_a)} ~~ and~~ by computation for function extensionality. Since$idtoiso\mathrm{isotoid}$~~ \idtoiso~~ isotoid ~~ are~~ and~~ inverses,~~~~ we~~~~ have~~$id\mathrm{idtoiso}idtoiso(\overline{\gamma}{)}_{a}{\gamma}_{a}$~~ \id{\idtoiso(\bar\gamma)_a}{\gamma_a}~~ idtoiso are inverses, we have ${idtoiso(\bar\gamma)_a}={\gamma_a}$ as desired.

Category theory functor category Rezk completion

category: category theory

Revision on September 5, 2018 at 13:23:49 by Ali Caglayan. See the history of this page for a list of all contributions to it.