functor precategory (Rev #8, changes)

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*This page is under construction. - Ali*

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 $B$. $\square$

We define a **natural isomorphism** to be an isomorphism? in $B^A$.

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 $idtoiso:({F}={G}) \to (F\cong G)$ is an equivalence.

To give an inverse to it, suppose $\gamma:F\cong G$ is a natural isomorphism. Then for any $a:A$, we have an isomorphism? $\gamma_a:F a \cong G a$, 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 ${F}={G}$ it will suffice to show that for any $a,b:A$, the functions

$F_{a,b}:hom_A(a,b) \to hom_B(F a,F b)$

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

become equal when transported along $\bar\gamma$. By computation for function extensionality, when applied to $a$, $\bar\gamma$ becomes equal to $isotoid(\gamma_a)$.

This reference needs to be included. For now as transports are not yet written up I didn’t bother including a reference to the page category. -Ali

But by~~ \cref{ct:idtoiso-trans},~~ [INCLUDE ME], 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 $(F\cong G) \to (F =G)$. Now consider the composite

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

Since hom-sets are sets?, their identity types are mere propositions?, so to show that two identities $p,q:F =G$ 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

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

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

Category theory functor category Rezk completion

category: category theory

Revision on September 6, 2018 at 06:55:38 by Ali Caglayan. See the history of this page for a list of all contributions to it.