## Natural transforms, products

Find two different functors $T : \textbf{Grp} \rightarrow \textbf{Grp}$ with the object function $T(G) = G$, the identity for every group $G$.

Solution: In other words, we need to find morphisms $f, g$, such that the corresponding morphisms $T f, T g$ are present in the image of the $T$. The two required morphisms are, given $x \in G$, $f(x) = x$, the identity morphism, and $g(x) = 1$, the zero morphism. Indeed, the functor carries identity to identity, and $T(f \circ g) = Tf \circ Tg$ $\Box$

Show how the field of quotients in an integral domain can be regarded as a functor.

Solution: An integral domain $G$ is defined as having no zero-divisors other than $0$. The field of quotients is a subgroup of this integral domain, where every element has an inverse. Indeed, we can define the functor $T : G \rightarrow Q$ to be the forgetful functor, which maps only units in $G$ to elements in $Q$. Let $f, g$ be two morphisms in $G$ for which $f \circ g$ is defined. Then, $T(1) = 1, T(f \circ g) = Tf \circ Tg$ $\Box$

Prove that there is no functor $T : \textbf{Grp} \rightarrow \textbf{Ab}$ sending each element of group $G$ to its `center`.

Proof: We have $a \in G$, and the object function $T(a) = bab^{-1}, b \in G$. Then, the object function takes identity to identity: $T(1) = bb^{-1} = 1$. But there is a problem with the arrow function: take $f = g = \text{id}$, so that $f \circ g = \text{id}$. Then, $(Tf \circ Tg)(a) = cbab^{-1}c^{-1} \neq T(f \circ g)(a) = bab^{-1}$ $\Box$

Let $S$ be a fixed set, and $X^S$ all functions $h : S \rightarrow X$. Show that $X \mapsto X^S$ is the object function of functor $\textbf{Set} \rightarrow \textbf{Set}$, and that evaluation $e_X : X^S \times X \overset{\bullet}{\rightarrow} X$ defined by $e(h, s) = h(s)$, the value of the function $h$ at $s \in S$, is an NT.

Proof: From the information given, we can conclude that $\mathscr{P} : X \mapsto X^S$ is the power-set functor, mapping every object $X \in \textbf{Set}$ to its power-set $\mathscr{P}X$. This is a functor, as identity maps to identity, and $\mathscr{P}(f \circ g) = \mathscr{P}f \circ \mathscr{P}g$ for any two morphisms $f, g \in \textbf{Set}$.

For the NT, the following diagram:

$$ \begin{xy} \xymatrix{ (\mathscr{P} \times X)b\ar[r]\ar[d] & \mathscr{P}(X)b\ar[d] \\ (\mathscr{P} \times X)b'\ar[r] & \mathscr{P}(X)b' } \end{xy} $$

commutes, for $b, b' \in X$ $\Box$

For a fixed group $H$, show that $G \mapsto H \times G$ defines a functor $H \times - : \textbf{Grp} \rightarrow \textbf{Grp}$, and that each morphism of groups $f : H \rightarrow K$ defines an NT $H \times - \overset{\bullet}{\rightarrow} K \times -$.

Proof: For the NT, the following diagram:

$$ \begin{xy} \xymatrix{ (H \times -)b\ar[r]\ar[d] & (K \times -)b\ar[d] \\ (H \times -)b'\ar[r] & (K \times -) b' } \end{xy} $$

commutes, for $b, b' \in X$.

Let $B, C$ be groups, and $S, T : B \rightarrow C$ functors (group homomorphisms). Then, show that an NT $S \overset{\bullet}{\rightarrow} T$ exists only if $S, T$ are `conjugate`; i.e. there is an element $h \in C$ such that $Tg = h(Sg)h^{-1}$, for all $g \in B$.

Find a category with an arrow that is both epi and monic, but not invertible.

Solution: $m : a \rightarrow b$ is such that $a \circ m = b \circ m \implies a = b$ and $m \circ a = m \circ b \implies a = b$, but $\exists$ no $m^{-1}$. This is true of an integral domain $\Box$

If $f \circ g$ is monic, so is $g$. Is it true of $f$?

Solution: The condition that $f \circ g$ is monic can be written as $(f \circ g) \circ a = (f \circ g) \circ b \implies a = b$, by cancelation of $f \circ g$. We know that $g$ is also monic, and this can be written as $g \circ a = g \circ b \implies a = b$. By comparing with the previous condition, we see that $f$ is also monic $\Box$

Prove that composites of monics is monic, and that of epis is epi.

Proof: This follows immediately from the previous problem $\Box$

Show that the product category includes the following special cases: product of monoids, groups, and sets.

Proof: A monoid is defined as set $M$ equipped with $\mu : M \times M \rightarrow M$ and $\eta : 1 \rightarrow M$, such that the following diagrams commute:

$$ \begin{xy} \xymatrix{ M \times M \times M \ar[r]^{1 \times \mu}\ar[d]^{\mu \times 1} & M \times M \ar[d]^{\mu} \\ M \times M \ar[r]^{\mu} & M } \end{xy} $$

$$ \begin{xy} \xymatrix{ 1 \times M \ar[r]^{\eta \times 1}\ar[d]^\lambda & M \times M \ar[d]^\mu & M \times 1 \ar[l]_{1 \times \eta}\ar[d]^\rho \\ M & M & M } \end{xy} $$

The product structure, written in terms of $M$ and $\mu$ is therefore:

$$ \begin{xy} \xymatrix{ & P\ar@{.>}[d]|{\exists!}\ar[dr]^f\ar[dl]_f & \\ M & M \times M\ar[l]_\mu\ar[r]^\mu & M } \end{xy} $$

Comparing with the first diagram for monoid, we get $P = M \times M \times M$ and $f = \mu \circ (1 \times \mu)$.

A group is defined as a monoid with an additional operation for inverse. We don't need this additional operation for the product structure to be defined.

$\textbf{Set}$ is simply a discrete category (category with one object), so the product structure is defined, in the obvious way, in terms of projections $\Box$

Show that the product of two preorders is a preorder.

Proof: A preorder is defined by reflexivity $a \leq a$, and transitivity $a \leq b \wedge b \leq c \implies a \leq c$. Let $a, b$ be objects and $f, g$ be arrows in the preorder category $C$. Then, $C \times C$ is defined as the category with objects as pairs $\langle a, b \rangle$ and arrows as pairs $\langle f, g \rangle$, where $f : a \rightarrow a', g : b \rightarrow b'$, such that:

$$ \langle a, b \rangle \overset{\langle f, g \rangle}{\longrightarrow} \langle a', b' \rangle $$

along with the usual product structure. Since $f, g$ act on $a, b$ independently, both $f$ and $g$ preserve preorder properties, and hence so does $\langle f, g \rangle$ $\Box$

Describe the opposite of the category $\textbf{Matr}_K$.

Solution: $\textbf{Matr}_K$ is the category of rectangular matrices of dimension $m \times n$ with entries in the commutative ring $K$. The objects are positive integers $m, n, \ldots$, and each $m \times n$ matrix $A$ is regarded as an arrow $A : n \rightarrow m$. The opposite category is a category with the same objects and direction of arrows reversed, i.e. the arrows are $A^{\text{op}} : m \rightarrow n$ $\Box$

## Functor categories, comma categories, free categories

Describe $B^X$, for $X$ a finite set.

Solution: $B^X$ is a category with objects the functors $T : \textbf{Set} \rightarrow B$, and arrows the morphisms between these functors. We can think of $T$ as functors that endow additional structure to the $\textbf{Set}$. Contained in $T$ is also the identity functor that identifies each element in $\textbf{Set}$ to a corresponding element in the richer category $B$ $\Box$

If $P, Q$ are preorders, describe $P^Q$, and show that it is a preorder.

Proof: The functor category may be written as $P^Q = \text{Funct}(Q, P)$, with objects the functors $T : Q \rightarrow P$, and morphisms the NTs between two such functors. A preorder is defined by reflexivity $a \leq a$ and transitivity $a \leq b \wedge b \leq c \implies a \leq c$; we need to show that $P^Q$ preserves this property.

Given categories $B, C$, and functor category $B^2$, show that each functor $H : C \rightarrow B^2$ determines the functors $S, T : C \rightarrow B$, and an NT $\tau : S \overset{\bullet}{\rightarrow} T$, and show that the assignment $H \mapsto \langle S, T, \tau \rangle$ is bijective.

Proof: $B^2$ is the category whose objects are arrows $f : a \rightarrow b \in B$, and whose arrows $f \rightarrow f'$ are those that, along with pair $\langle h, k \rangle$, make the following diagram commute:

$$ \begin{xy} \xymatrix{ a \ar[r]^h\ar[d]^f & a' \ar[d]^{f'} \\ b \ar[r]^k & b' } \end{xy} $$

For small categories $A, B, C$, establish the bijection: $$\textbf{Cat}(A \times B, C) \cong \textbf{Cat}(A, C^B)$$

and show that it is natural in $A, B, C$.

Show that the horizontal composition $\circ$ is a functor: $$\circ : A^B \times B^C \rightarrow A^C$$

Let $S$ be a set with two everywhere-defined binary operations $\bullet : S \times S \rightarrow S$ and $\circ : S \times S \rightarrow S$, such that both have the same two-sided identity element $e$, and satisfy the interchange law:

$$ (\tau' \bullet \sigma') \circ (\tau \bullet \sigma) = (\tau \circ \tau') \bullet (\sigma \circ \sigma') $$

Prove that $\circ$ and $\bullet$ are equal, and that each is commutative.

If $K$ is a commutative ring, show that the [comma category](/category/1#comma-category) $(K \downarrow \textbf{CRng})$ is the category of small commutative [$K$-algebras](/ac/1#algebras).

Solution: