[ct/hw] Homework


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](/ct/1#comma-category) \((K \downarrow \textbf{CRng})\) is the category of small commutative [\(K\)-algebras](/ac/1#algebras).

Solution: