Homework 1

Problem 1

An object $i$ in a category $C$ is called initial if for every object $c \in C$ there exists a unique arrow $i \to c$ ; in other words, the hom-sets ${Hom}_{C} (i, c)$ are all singleton sets.

Dually, an object $t$ is called terminal if for every object $c \in C$ there exists a unique arrow $c \to t$ , i.e., the hom-sets ${Hom}_{C} (c, t)$ are all singleton sets.

Show that an initial (respectively, terminal) object, if it exists, is unique up to unique isomorphism.
Show that in $S e t$ , the empty set is initial and the singleton set is terminal.
Show that in $A b$ , the trivial group $0$ is both initial and terminal. (Such an object is called a null or zero object.)
Show that $F i e l d$ contains neither an initial nor a terminal object.

Solution

Suppose $i, i^{'} \in C$ are two initial objects. Since $i$ is initial, there is a unique arrow $f : i \to i^{'}$ in $C$ ; since $i^{'}$ is initial, there is a unique arrow $g : i^{'} \to i$ in $C$ . We claim $f$ and $g$ are mutual inverses (hence isomorphisms). To that end, observe that $g \circ f : i \to i$ is an arrow from $i$ to itself. However, another such arrow from $i$ to itself is the identity arrow $1_{i} : i \to i$ . As $i$ is initial, there can only exist one arrow from $i$ to itself, hence we must have $g \circ f = 1_{i}$ . Similarly, the composition $f \circ g$ must equal the identity arrow $1_{i^{'}}$ .

The "dual" argument for terminal objects proceeds similarly. Suppose $t, t^{'} \in C$ are both terminal objects. Since $t$ is terminal, there is a unique arrow $f : t^{'} \to t$ ; since $t^{'}$ is terminal, there is a unique arrow $g : t \to t^{'}$ . The composition $g \circ f$ is an arrow from $t^{'}$ to itself, hence (by uniqueness) must equal the identity arrow $1_{t^{'}}$ . Similarly, the composition $f \circ g$ must equal the identity arrow $1_{t}$ , so $f$ and $g$ are mutual inverses (and hence both isomorphisms).
For every set $X$ there is a unique set map $\emptyset \to X$ , namely the so-called "empty map". Similarly, for each fixed singleton set ${⋆}$ there is a unique set map $X \to {⋆}$ , namely the map that sends every element of $X$ to $⋆$ .
Since group morphisms must preserve identities, to each abelian group $A$ there is a unique group morphism $0 \to A$ , namely $0 \mapsto 0_{A}$ . And just as with sets, there is a unique group morphism $A \to 0$ , namely the one that maps every element of $A$ to $0$ . (What happened to the empty set? Remember: groups are required to have an identity element, and hence must be nonempty!)
Recall the following fact: there are no morphisms between fields of different characteristics. In particular, there does exist a field $F$ that has morphisms to both $F_{2}$ and $F_{3}$ , say, and hence there does not exist a field with morphisms to every field (let along unique morphisms). Similarly, there does not exist a field with morphisms from every field.

Problem 2

An arrow $f : a \to b$ in a category $C$ is called monic (or a monomorphism) if it is left-cancellable; i.e., if whenever $g_{1}, g_{2} : c \to a$ are arrows such that $f \circ g_{1} = f \circ g_{2}$ , then $g_{1} = g_{2}$ .

Dually, the arrow $f$ is called epic (or epi, or an epimorphism) if it is right-cancellable; i.e., if whenever $g_{1}, g_{2} : b \to c$ are arrows such that $g_{1} \circ f = g_{2} \circ f$ , then $g_{1} = g_{2}$ .

In $S e t$ , show that a set map $f : X \to Y$ is monic (respectively, epic) if and only if it is injective (respectively, surjective).
Show that, in a general category $C$ , if an arrow $f : a \to b$ is an isomorphism (i.e., invertible), then $f$ is both monic and epic.
Show that in $R i n g$ , the ring inclusion $i : Z \to Q$ is both monic and epic, even though the map $i$ is not surjective.

Solution

First suppose $f : X \to Y$ is monic. Suppose $x_{1}, x_{2} \in X$ are elements such that $f (x_{1}) = f (x_{2})$ . Define set maps $g_{1}, g_{2} : {⋆} \to X$ by $g_{i} (⋆) = x_{i}$ . Then $(f \circ g_{1}) (⋆) = f (x_{1}) = f (x_{2}) = (f \circ g_{2}) (⋆)$ , so $f \circ g_{1} = f \circ g_{2}$ . Since $f$ is monic, this implies $g_{1} = g_{2}$ and hence $x_{1} = g_{1} (⋆) = g_{2} (⋆) = x_{2}$ . Thus, $f$ is indeed injective.

Conversely, suppose $f : X \to Y$ is injective. Suppose $g_{1}, g_{2} : Z \to X$ are set maps such that $f \circ g_{1} = f \circ g_{2}$ . Take any element $z \in Z$ . Then $f (g_{1} (z)) = f (g_{2} (z))$ , so by the injectivity of $f$ we must have $g_{1} (z) = g_{2} (z)$ . Since $z \in Z$ was arbitrary, this proves $g_{1} = g_{2}$ . Thus, $f$ is left cancellable, i.e., monic.

Next suppose $f : X \to Y$ is epic and suppose, towards a contradiction, that $f$ is not surjective. Then we can choose some element $y_{0} \in Y$ that is not in the image of $X$ . Now consider two set maps $g_{1}, g_{2} : Y \to {0, 1}$ . For the first map, simply define $g_{1} (y) = 1$ for all $y \in Y$ . For the second map, define $g_{2}$ to be the characteristic function of the image $f (X) \subseteq Y$ ; in other words, set $g (y) = 1$ if there exists some $x \in X$ with $f (x) = y$ , and set $g_{2} (y) = 0$ otherwise. Now notice that for every $x \in X$ we have $(g_{1} \circ f) (x) = 1$ (since $g_{1}$ takes the constant value $1$ ), and also $(g_{2} \circ f) (x) = g_{2} (f (x)) = 1$ (since clearly $f (x)$ is in the image of $f$ ). Thus, $g_{1} \circ f = g_{2} \circ f$ . Since $f$ is epic, this implies $g_{1} = g_{2}$ . But this is a contradiction, since $g_{1} (y_{0}) = 1 \neq 0 = g_{2} (y_{0})$ .

Finally, suppose $f : X \to Y$ is surjective. Suppose $g_{1}, g_{2} : Y \to Z$ are two set maps such that $g_{1} \circ f = g_{2} \circ f$ . Take any element $y \in Y$ . Since $f$ is surjective, there exists some $x \in X$ with $f (x) = y$ . We then have $g_{1} (y) = g_{1} (f (x)) = (g_{1} \circ f) (x) = (g_{2} \circ f) (x) = g_{2} (y)$ . Since $y \in Y$ was arbitrary, this proves $g_{1} = g_{2}$ ; i.e., $f$ is epic.
Suppose an arrow $f : a \to b$ in a category $C$ is an isomorphism. Let $g : b \to a$ be the inverse arrow. To show $f$ is monic, suppose $h_{1}, h_{2} : c \to a$ are two arrows such that $f \circ h_{1} = f \circ h_{2}$ . Composing both sides with $g$ then gives $g \circ (f \circ h_{1}) = g \circ (f \circ h_{2})$ . But now notice that

$g \circ (f \circ h_{1}) = (g \circ f) \circ h_{1} = 1_{a} \circ h_{1} = h_{1} .$

Similarly, $g \circ (f \circ h_{2}) = h_{2}$ . We therefore have $h_{1} = h_{2}$ , and so $f$ is indeed monic.

The argument that $f$ is epic is essentially the same, but we provide here for reference. Suppose $h_{1}, h_{2} : b \to c$ are two arrows such that $h_{1} \circ f = h_{2} \circ f$ . Precomposing both sides by $g$ gives $(h_{1} \circ f) \circ g = (h_{2} \circ f) \circ g$ . As before, associativity then implies $h_{1} \circ 1_{b} = h_{2} \circ 1_{b}$ and hence $h_{1} = h_{2}$ . Thus, $f$ is epic.
First recall that the inclusion $i : Z \to Q$ simply sends each integer $n$ to the rational number $\frac{n}{1}$ . In particular, strictly at the level of set maps this is injective. Now suppose $g_{1}, g_{2} : R \to Z$ are ring morphisms such that $i \circ g_{1} = i \circ g_{2}$ as ring morphisms $R \to Q$ . Take any $r i n R$ and consider the integers $n_{1} = g_{1} (r)$ and $n_{2} = g_{2} (r)$ . We have $i (n_{1}) = i (g_{1} (r)) = (i \circ g_{1}) (r) = (i \circ g_{2}) (r) = i (g_{2} (r)) = i (n_{2})$ . Since $i$ is injective (as a set map), this implies $n_{1} = n_{2}$ , i.e., $g_{1} (r) = g_{2} (r)$ . Since $r \in R$ was arbitrary, this proves $g_{1} = g_{2}$ . Thus, $i$ is monic.

The fact that $i$ is epic is more surprising and essentially boils down to the fact that ring morphisms from $Q$ are entirely determined by their restriction to $Z$ . To see this in detail, suppose $g_{1}, g_{2} : Q \to R$ are ring morphisms such that $g_{1} \circ i = g_{2} \circ i$ as ring morphisms $Z \to R$ . Take any element in $Q$ and write it in the form $\frac{m}{n}$ with $m, n \in Z$ . Since each $g_{i} : Q \to R$ is a ring morphism, we must have

$g_{i} (\frac{m}{n}) = g_{i} (\frac{m}{1} \cdot \frac{1}{n}) = g_{i} (\frac{m}{1}) g_{i} {(\frac{n}{1})}^{- 1} .$

Now note that $g_{1} (\frac{m}{1}) = g_{1} (i (m)) = (g_{1} \circ i) (m) = (g_{2} \circ i) (m) = g_{2} (\frac{m}{1}) .$ Combined with the above equality, it now immediately follows that $g_{1} (\frac{m}{n}) = g_{2} (\frac{m}{n})$ . Since the rational number $\frac{m}{n}$ was arbitrary, this proves $g_{1} = g_{2}$ ; i.e., $i$ is epic.

Problem 3

There are two functors vying for the name "power set functor".

For the first, define $P : S e t \to S e t$ as follows:

On objects: for each set $X$ , define $P (X)$ to be the power set of $X$ , i.e., the set of all subsets of $X$ .
On arrows: for each set map $f : X \to Y$ , define a set map $P (f) : P (X) \to P (Y)$ by sending each subset $S \subseteq X$ to its image $f (S) \subseteq Y$ .

For the second, define $P^{'} : {S e t}^{op} \to S e t$ as follows^[1]:

On objects: for each set $X$ , define $P^{'} (X)$ to once again be the power set of $X$ .
On arrows: for each arrow $f^{op} : X \to Y$ (corresponding to a set map $f : Y \to X$ ), define a set map $P^{'} (f^{op}) : P^{'} (X) \to P^{'} (Y)$ by sending each subset $S \subseteq X$ to its preimage $f^{- 1} (X) \subset Y$ . (Recall that $f^{- 1} (X) = {y \in Y ∣ f (y) \in X}$ .)

Verify that $P$ and $P^{'}$ are indeed both functors. The first is sometimes called the covariant power set functor and the second the contravariant power set functor.

Solution

Let's first verify $P : S e t \to S e t$ is a functor. Fix any set $X$ and let $1_{X} : X \to X$ be the identity set map. Then every subset $S \subseteq X$ the identity set map $1_{X}$ on $X$ sends $S$ to itself, i.e., $1_{X} (S) = S$ . Thus, $1_{X}$ induces the identity map on $P (X)$ , i.e., $P (1_{X}) = 1_{P (X)}$ .

Next suppose $f : X \to Y$ and $g : Y \to Z$ are composable set maps. We must show $P (g \circ f) = P (g) \circ P (f)$ as set maps $P (X) \to P (Z)$ . To that end, fix any $S \in P (X)$ , i.e., any subset $S \subseteq X$ . It follows from the very definition of composition for set maps that $g (f (S)) = (g \circ f) (S)$ . This exactly implies $P (g) (P (f) (S)) = P (g \circ f) (S)$ . Since $S \in P (X)$ was arbitrary, this proves $P (g) \circ P (f) = P (g \circ f)$ .

Let's similarly prove $P^{'} : {S e t}^{op} \to S e t$ is a functor. Once again, first fix a set $X$ and let $1_{X} : X \to X$ be the identity set map. Then for every subset $T \subseteq X$ we certainly have $1_{X}^{- 1} (T) = T$ , and so $P^{'}$ induces the identity map on $P^{'} (X)$ .

Finally, suppose $f^{op} : X \to Y$ and $g^{op} : Y \to Z$ are composable arrows in ${S e t}^{op}$ . Note that these arrows (by the definition of the opposite category ${S e t}^{op})$ correspond to set maps $f : Y \to X$ and $g : Z \to Y$ , and also that $g^{op} \circ f^{op} = (f \circ g)^{op} : X \to Z$ . Now fix any element $S \in P^{'} (X)$ , i.e., subset $S \subseteq T$ . Then we have

$\begin{aligned} P^{'} (g^{op} \circ f^{op}) (S) & = P^{'} ((f \circ g)^{op}) (S) \\ = (f \circ g)^{- 1} (S) \\ = g^{- 1} (f^{- 1} (S)) \\ = P^{'} (g^{op}) (P^{'} (f^{op} (S)) \\ (P^{'} (g^{op}) \circ P^{'} (f^{op})) (S) . \end{aligned}$

Here we used the easy fact from basic set theory that $(f \circ g)^{- 1} (S) = g^{- 1} (f^{- 1} (S))$ , which does feel a bit like cheating. In any case, since $S \in P^{'} (X)$ was arbitrary, this does indeed prove $P^{'} (g^{op} \circ f^{op} P^{'} (g^{op}) \circ P^{'} (f^{op} .$ Thus, $P^{'}$ is indeed a functor.

Hate all of the "opposite" notation?

If you really hated all of the opposite notation used above, you could have instead proved $P^{'}$ was a "contravariant" functor and saved yourself some notational awkwardness.

Problem 4

Prove there does not exist a functor $G r p \to A b$ with object function sending each group $G$ to its center.

Hint

Consider a certain sequence of group morphisms $S_{2} \to S_{3} \to S_{2}$ .

Solution

We will show that the assignment to each group $G$ its center $Z (G)$ is not functorial, by finding a pair of composable morphisms $f, g$ such that $Z (g \circ f)$ cannot possibly equal $Z (g) \circ Z (f)$ .

Following the hint, consider the morphism $p : S_{3} \to S_{2}$ given by composition of the projection morphism $S_{3} \to S_{3} / A_{3}$ with the isomorphism $S_{3} / A_{3} ≅ S_{2}$ . One can readily verify that the composition of this morphism $p : S_{3} \to S_{2}$ with the inclusion morphism $i : S_{2} \to S_{3}$ gives the identity morphism $S_{2} \to S_{2}$ .

Now suppose there were some functor $Z : Grp \to Ab$ that mapped each group $G$ to its center $Z (G)$ . Since $Z (S_{2}) = S_{2}$ and $Z (S_{3}) = {e}$ , the functor $Z$ must map the morphism $i : S_{2} \to S_{3}$ to the trivial morphism $Z (i) : S_{2} \to {e}$ . This forces the composition $Z (p) \circ Z (i)$ to be the trivial morphism $S_{2} \to S_{2}$ that maps everything to the identity. On the other hand, since $p \circ i$ is the identity morphism on $S_{2}$ , by the assumption that $Z$ is a functor we must have that $Z (p \circ i)$ is the identity morphism on $Z (S_{2}) = S_{2}$ . Thus, we cannot have $Z (p) \circ Z (i) = Z (p \circ i)$ , contradicting the assumption that $Z$ was a functor (and hence must be compatible with composition).

Another way to say this all very succinctly is that group morphisms don't respect membership in centers. Our example is that of the element $(1, 2)$ that is in the center of $S_{2}$ but maps (under the usual inclusion of $S_{2}$ into $S_{3}$ ) to an element not in the center of $S_{3}$ .

Here we are using the opposite category to avoid talking about contravariant functors. ↩︎