nLab
cartesian closed category

Definition

A cartesian closed category (sometimes: ccc) is a category with finite products which is closed with respect to its cartesian monoidal structure.

The internal hom $[S,X]$ in a cartesian closed category is often called exponentiation and is denoted $X^S$.

A cartesian closed functor between cartesian closed categories $C$, $D$ is a functor $F:C\to D$ that is product-preserving and that is also exponential-preserving, meaning that the canonical map

$$F(a^b)\to F(a{)}^{F(b)},$$F(a^b) \to F(a)^{F(b)},

corresponding to the composite

$$F(a^b)\times F(b)\cong F(a^b\times b)\stackrel{F({\mathrm{eval}}_{a,b})}{\to }F(a),$$F(a^b) \times F(b) \cong F(a^b \times b) \stackrel{F(eval_{a, b})}{\to} F(a),

is an isomorphism.

Examples

• Any topos or quasitopos, such as Set, is cartesian closed.

  See also closed monoidal structure on presheaves.

• Cat is also cartesian closed.

• Many nice categories of topological spaces are also cartesian closed, particularly the convenient categories of spaces.

Some basic consequences

A category is cartesian closed if it has finite products and if for any two objects $X$, $Y$, there is an object $Y^X$ (thought of as a “space of maps from $X$ to $Y$”) such that for any object $Z$, there is a bijection between the set of maps $Z\to Y^X$ and the set of maps $Z\times X\to Y$, and this bijection is natural in $Z$.

There is an evaluation map ${\mathrm{ev}}_{X,Y}:Y^X\times X\to Y$, which by definition is the map corresponding to the identity map $Y^X\to Y^X$ under the bijection. Using the naturality, it may be shown that the bijection $\mathrm{hom}(Z,Y^X)\to \mathrm{hom}(Z\times X,Y)$ takes a map $\varphi :Z\to Y^X$ to the composite

$$Z\times X\stackrel{\varphi \times 1_X}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }Y$$Z \times X \stackrel{\phi \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y

and the universal property of $Y^X$ may be phrased thus: given any $\psi :Z\times X\to Y$, there exists a unique map $\varphi :Z\to Y^X$ for which $\psi ={\mathrm{ev}}_{X,Y}\circ (\varphi \times 1_X)$.

Taking $Z=1$ (the terminal object), maps $1\to Y^X$ (or “points” of $Y^X$) are in bijection with maps $X\stackrel{{\pi }_2^{-1}}{\to }1\times X\to Y$. So the “underlying set” of $Y^X$, namely $\mathrm{hom}(1,Y^X)$, is identified with the set of maps from $X$ to $Y$. Let us denote the point corresponding to $f:X\to Y$ by $[f]:1\to Y^X$. Then, by definition,

$$(1\times X\stackrel{[f]\times 1_X}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }Y)=(1\times X\stackrel{{\pi }_2}{\to }X\stackrel{f}{\to }Y)$$(1 \times X \stackrel{[f] \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y) = (1 \times X \stackrel{\pi_2}{\to} X \stackrel{f}{\to} Y)

The following lemma says that the internal evaluation map ${\mathrm{ev}}_{X,Y}$ indeed behaves as an evaluation map at the level of underlying sets.

Lemma

Given a map $f:X\to Y$ and a point $x:1\to X$, the composite

$$1\stackrel{⟨[f],x⟩}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }Y$$1 \stackrel{\langle [f], x \rangle}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y

is the point $fx:1\to Y$.

Proof

We have

$$\begin{array}{ccc}1\stackrel{⟨[f],x⟩}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }X& =& 1\stackrel{{\delta }_1}{\to }1\times 1\stackrel{[f]\times x}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }Y\\ & =& 1\stackrel{{\delta }_1}{\to }1\times 1\stackrel{1\times x}{\to }1\times X\stackrel{[f]\times 1_X}{\to }{Y}^X\times X\stackrel{{\mathrm{ev}}_{X,Y}}{\to }Y\\ & =& 1\stackrel{{\delta }_1}{\to }1\times 1\stackrel{1\times x}{\to }1\times X\stackrel{{\pi }_2}{\to }X\stackrel{f}{\to }Y\\ & =& 1\stackrel{{\delta }_1}{\to }1\times 1\stackrel{{\pi }_2}{\to }1\stackrel{x}{\to }X\stackrel{f}{\to }Y\\ & =& 1\stackrel{x}{\to }X\stackrel{f}{\to }Y\end{array}$$\array{ 1 \stackrel{\langle [f], x \rangle}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} X & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{[f] \times x}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{1 \times x}{\to} 1 \times X \stackrel{[f] \times 1_X}{\to} Y^X \times X \stackrel{ev_{X, Y}}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{1 \times x}{\to} 1 \times X \stackrel{\pi_2}{\to} X \stackrel{f}{\to} Y\\ & = & 1 \stackrel{\delta_1}{\to} 1 \times 1 \stackrel{\pi_2}{\to} 1 \stackrel{x}{\to} X \stackrel{f}{\to} Y\\ & = & 1 \stackrel{x}{\to} X \stackrel{f}{\to} Y }

where the penultimate equation uses naturality of the projection map ${\pi }_2$.

Definition

Internal composition $c_{X,Y,Z}:Z^Y\times Y^X\to Z^X$ is the unique map such that

$$(Z^Y\times Y^X\times X\stackrel{c\times 1_X}{\to }{Z}^X\times X\stackrel{{\mathrm{ev}}_{X,Z}}{\to }Z)=(Z^Y\times Y^X\times X\stackrel{1\times {\mathrm{ev}}_{X,Y}}{\to }{Z}^Y\times Y\stackrel{{\mathrm{ev}}_{Y,Z}}{\to }Z)$$(Z^Y \times Y^X \times X \stackrel{c \times 1_X}{\to} Z^X \times X \stackrel{ev_{X, Z}}{\to} Z) = (Z^Y \times Y^X \times X \stackrel{1 \times ev_{X, Y}}{\to} Z^Y \times Y \stackrel{ev_{Y, Z}}{\to} Z)

One may show that internal composition behaves as the usual composition at the underlying set level, in that given maps $g:Y\to Z$, $f:X\to Y$, we have

$$(1\stackrel{⟨[g],[f]⟩}{\to }{Z}^Y\times Y^X\stackrel{c_{X,Y,Z}}{\to }{Z}^X)=[gf]:1\to Z^X$$(1 \stackrel{\langle [g], [f] \rangle}{\to} Z^Y \times Y^X \stackrel{c_{X, Y, Z}}{\to} Z^X) = [g f]: 1 \to Z^X

Properties

• In a cartesian closed category, the product functors $A\times -$ have right adjoints, so they preserve all colimits. In particular, a cartesian closed category that has finite coproducts is a distributive category.

• The internal logic of cartesian closed categories is the typed lambda-calculus.

Inheritance by reflective subcategories

In showing that a given category is cartesian closed, the following theorem is often useful (cf. A4.3.1 in the Elephant):

Theorem

If $C$ is cartesian closed, and $D\subseteq C$ is a reflective subcategory, then the reflector $L:C\to D$ preserves finite products if and only if $D$ is an exponential ideal (i.e. $Y\in D$ implies $Y^X\in D$ for any $X\in C$). In particular, if $L$ preserves finite products, then $D$ is cartesian closed.

Exponentials of cartesian closed categories

The following observation was taken from a post of Mike Shulman at MathOverflow.

If $𝒞$ is small and $𝒟$ is complete and cartesian closed, then ${𝒟}^{𝒞}$ is also complete and cartesian closed. Completeness is clear since limits in $D^C$ are computed pointwise. As for cartesian closure, we can compute exponentials in essentially the same way as for presheaves, motivated by $𝒟$-enriched category theory:

$$G^F(x)={\int }_{y\in 𝒞}\prod _{𝒞(x,y)}G(y{)}^{F(y)}.$$G^F(x) = \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} G(y)^{F(y)}.

Then we can compute

$$\begin{array}{ccc}{𝒟}^{𝒞}(H,G^F)& =& {\int }_{x\in 𝒞}𝒟(H(x),G^F(x))\\ & =& {\int }_{x\in 𝒞}𝒟(H(x),{\int }_{y\in 𝒞}\prod _{𝒞(x,y)}G(y{)}^{F(y)})\\ & =& {\int }_{x\in 𝒞}{\int }_{y\in 𝒞}\prod _{𝒞(x,y)}𝒟(H(x),G(y{)}^{F(y)})\\ & =& {\int }_{y\in 𝒞}𝒟({\int }^{x\in 𝒞}\sum _{𝒞(x,y)}H(x),G(y{)}^{F(y)})\\ & =& {\int }_{y\in 𝒞}𝒟(H(y),G(y{)}^{F(y)})\\ & =& {\int }_{y\in 𝒞}𝒟(H(y)\times F(y),G(y))\\ & =& {𝒟}^{𝒞}(H\times F,G).\end{array}$$\array{ \mathcal{D}^{\mathcal{C}}\left(H,G^F\right) &=& \int_{x\in \mathcal{C}} \mathcal{D}\left(H(x), G^F(x)\right)\\ &=& \int_{x\in \mathcal{C}} \mathcal{D}\left(H(x), \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} G(y)^{F(y)}\right)\\ &=& \int_{x\in \mathcal{C}} \int_{y\in \mathcal{C}} \prod_{\mathcal{C}(x,y)} \mathcal{D}\left(H(x), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left( \int^{x\in \mathcal{C}} \sum_{\mathcal{C}(x,y)} H(x), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left(H(y), G(y)^{F(y)}\right)\\ &=& \int_{y\in \mathcal{C}} \mathcal{D}\left(H(y)\times F(y), G(y)\right)\\ &=& \mathcal{D}^{\mathcal{C}}(H\times F, G). }

Here the antepenultimate step uses the co-Yoneda lemma. This appears to involve colimits in $𝒟$ as well, but the existence of these colimits is not actually an extra assumption: the co-Yoneda lemma tells us that ${\int }^{x\in 𝒞}{\sum }_{𝒞(x,y)}H(x)$ exists and is isomorphic to $H(y)$.

Similarly, the above argument can be interpreted to say that even if $𝒟$ is not complete, then the exponential $G^F$ in ${𝒟}^{𝒞}$ exists if and only if the particular limits above exist, and in that case they are isomorphic.

A more abstract argument using comonadicity and the adjoint triangle theorem, which also applies to locally cartesian closed categories, can be found in Theorem 2.12 of

• Street and Verity, The comprehensive factorization and torsors, TAC

and is reproduced in the setting of closed monoidal categories at closed monoidal category.

Functional completeness theorem

Theorem 1

Let $C$ be a cartesian closed category, and $c$ an object of $C$. Then the Kleisli category of the comonad $c\times -:C\to C$, denoted $C[c]$, is also a cartesian closed category.

Remark: The Kleisli category of the comonad $c\times -$ on $C$ is canonically equivalent to the Kleisli category of the monad $(-{)}^c$ on $C$.

Proof of Theorem 1

Let $a$ and $b$ be objects of $C[c]$. Claim: the product $a\times b$ in $C$, considered as an object of $C[c]$, is the product of $a$ and $b$ in $C[c]$, according to the following series of natural bijections:

$$\begin{array}{cc}\underline{z\to az\to b}& C[c]\\ \underline{c\times z\to a,c\times z\to b}& C\\ \underline{c\times z\to a\times b}& C\\ z\to a\times b& C[c]\end{array}$$\array{ \underline{z \to a \qquad z \to b} & C[c] \\ \underline{c \times z \to a, c \times z \to b} & C \\ \underline{c \times z \to a \times b} & C \\ z \to a \times b & C[c] }

Claim: the exponential $a^b$ in $C$, considered as an object of $C[c]$, is the exponential of $a$ and $b$ in $C[c]$, according to the following series of natural bijections:

$$\begin{array}{cc}\underline{z\to a^b}& C[c]\\ \underline{c\times z\to a^b}& C\\ \underline{c\times z\times b\to a}& C\\ z\times b\to a& C[c]\end{array}$$\array{ \underline{z \to a^b} & C[c] \\ \underline{c \times z \to a^b} & C \\ \underline{c \times z \times b \to a} & C \\ z \times b \to a & C[c] }

where in the last step, we used the prior claim that the product of objects in $C$, when viewed as an object of $C[c]$, gives the product in $C[c]$.

Observe that the object $c$ in $C[c]$ has an element $e:1\to c$, corresponding to the canonical isomorphism $c\times 1\cong c$ in $C$. We call this element the “generic element” of $c$ in $C[c]$, according to the following theorem. This theorem can be viewed as saying that in the doctrine of cartesian closed categories, $C[c]$ is the result of freely adjoining an arrow $1\to c$ to $C$.

Theorem (Functional completeness)

Let $C$ and $D$ be cartesian closed categories, and let $F:C\to D$ be a cartesian closed functor. Let $c$ be an object of $C$, and let $t:F(1)\to F(c)$ be an element in $D$. Then there exists an extension of $F$ to a cartesian closed functor $\tilde{F}:C[c]\to D$ that takes the generic element $e:1\to c$ in $C[c]$ to the element $t$, and this extension is unique up to isomorphism.

Proof (sketch)

On objects, $\tilde{F}(a)=F(a)$. Let $f:a\to b$ be a map of $C[c]$, i.e., let $g:c\times a\to b$ be the corresponding map in $C$, and define $\tilde{F}(f)$ to be the composite

$$F(a)\cong 1\times F(a)\stackrel{t\times 1}{\to }F(c)\times F(a)\cong F(c\times a)\stackrel{F(g)}{\to }F(b)$$F(a) \cong 1 \times F(a) \stackrel{t \times 1}{\to} F(c) \times F(a) \cong F(c \times a) \stackrel{F(g)}{\to} F(b)

It is straightforward to check that $\tilde{F}$ is a cartesian closed functor and is, up to isomorphism, the unique cartesian closed functor taking $e$ to $t$.

Related concepts

• cartesian closed category, locally cartesian closed category

• cartesian closed functor, locally cartesian closed functor

• cartesian closed model category, locally cartesian closed model category

• cartesian closed (∞,1)-category locally cartesian closed (∞,1)-category