[[!redirects natural isomorphism]] +--{.query} *This page is under construction. - Ali* =-- * table of contents {:toc} ## Definition ## 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 $B^A$. ## Properties ## ### Lemma 9.2.4 ### 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 $\delta \gamma 1_F.\ \square$ ### Theorem 9.2.5 ### 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 $isotoid(\gamma_a)$. +--{.query} 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 [**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. $\square$ ## See also ## [[Category theory]] [[functor]] [[category]] [[Rezk completion]] ## References ## [[HoTT Book]] category: category theory