Session 23: More on universal mapping properties, Part 2

Section 5.18 Session 23: More on universal mapping properties, Part 2

Let’s pick up with the second half of Exercise 1:

Example 5.18.1. Exercise 1: (Part 2/?).

Continued from last week...

Solution.

Let’s begin with reconstructing the diagram:

$Structure of maps in $\mathcal{C}_{B_1 B_2}$$

The goal here is use these objects $B_1$ and $B_2$ in $\mathcal{C}$ as the basis for constructing a new category $\mathcal{C}_{B_1 B_2}\text{.}$

We define an object in $\mathcal{C}_{B_1 B_2}$ as an object of $\mathcal{C}$ equiped with a pair of maps to $B_1$ and $B_2\text{.}$

We define a map in $\mathcal{C}_{B_1 B_2}$ as a map in $\mathcal{C}$ which preserves the structure of the commutative diagram above. Specifically, any map $X \xrightarrow{f} Y$ in $\mathcal{C}_{B_1 B_2}$ must satisfy the relations $\psi_1 f = \phi_1$ and $\psi_2 f = \phi_2\text{.}$

Since every $\mathcal{C}_{B_1 B_2}$-map comes from a map $\mathcal{C}\text{,}$ it must have a defined domain and codomain. Likewise, since the objects in $\mathcal{C}_{B_1 B_2}$ come from objects in $\mathcal{C}$ they must each have an identity map. For any object $X$ in $\mathcal{C}_{B_1 B_2}\text{,}$ a simple substitution $\phi_1 1_X = \phi_1$ and $\phi_2 1_X = \phi_2$ shows the identity maps from $\mathcal{C}$ preserve structure.

Let’s take a second to verify that $\mathcal{C}_{B_1 B_2}$ is closed under composition. Consider two maps $X \xrightarrow{f} Y \xrightarrow{g} Z\text{.}$

$Composition in $\mathcal{C}_{B_1 B_2}$$

We can assume the following relations from the definition of $\mathcal{C}_{B_1 B_2}$-map:

\begin{equation*} \psi_1 f = \phi_1 \end{equation*}

\begin{equation*} \psi_2 f = \phi_2 \end{equation*}

\begin{equation*} \chi_1 g = \psi_1 \end{equation*}

\begin{equation*} \chi_2 g = \psi_2 \end{equation*}

In order for $g \circ f$ to be a valid $\mathcal{C}_{B_1 B_2}$-map, we need to establish both $\chi_1 (g \circ f) = \phi_1$ and $\chi_1 (g \circ f) = \phi_1\text{.}$ This follows easily from the associative property in $\mathcal{C}$ and substitution with the relations above:

\begin{equation*} \chi_1 (g \circ f) = (\chi_1 \circ g) \circ f = \psi_1 \circ f = \phi_1 \end{equation*}

\begin{equation*} \chi_2 (g \circ f) = (\chi_2 \circ g) \circ f = \psi_2 \circ f = \phi_2 \end{equation*}

The only thing left to prove is the associative law. Let’s consider a set of three $\mathcal{C}_{B_1 B_2}$-maps $A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D\text{.}$ Each of these maps needs to correspond with a pair of maps to $B_1$ and $B_2\text{,}$ as per the following diagram:

$Associative property in $\mathcal{C}_{B_1 B_2}$$

By our definition of composition above, we know we have $g \circ f$ satisfying $\chi_1 (g \circ f) = \phi_1$ and $\chi_2 (g \circ f) = \phi_2\text{.}$ Postcompose with $h$ to get $h \circ (g \circ f)$ with $\omega_1 (h \circ (g \circ f)) = \chi_1 (g \circ f) = \phi_1$ and $\omega_1 (h \circ (g \circ f)) = \chi_1 (g \circ f) = \phi_1\text{.}$

Similarly, we must have $h \circ g$ satisfying $\omega_1 (h \circ g) = \psi_1$ and $\omega_2 (h \circ g) = \psi_2\text{.}$ Precompose with $f$ to get $(h \circ g) \circ f$ with $\omega_1 ((h \circ g) \circ f) = (\omega_1 (h \circ g)) \circ f = \psi_1 f = \phi_1$ and $\omega_2 ((h \circ g) \circ f) = (\omega_2 (h \circ g)) \circ f = \psi_2 f = \phi_2\text{.}$

Since we know that the associative property holds in $\mathcal{C}$ and have shown that it satisfies the structure preservation of $\mathcal{C}_{B_1 B_2}\text{,}$ that should complete the proof that $\mathcal{C}_{B_1 B_2}$ is a valid category.

Knowing $\mathcal{C}_{B_1 B_2}$ is a category, what should the terminal object look like? Per definition, a terminal object $T$ should have the property that for every object $X$ in $\mathcal{C}_{B_1 B_2}\text{,}$ there is exactly one map $X \xrightarrow{\tau} T\text{.}$ As an object in $\mathcal{C}_{B_1 B_2}\text{,}$ the terminal object $T$ must be accompanied by a pair of maps $T \xrightarrow{\tau_1} B_1$ and $T \xrightarrow{\tau_2} B_1$ such that $\tau_1 \tau = \phi_1$ and $\tau_2 \tau = \phi_2$ where $\phi_1,\phi_2$ are the respective pair of maps associated with $X\text{.}$

Since every object has an identity map, this rule must also apply to $T\text{.}$ Since any $X \rightarrow T$ is unique, then the uniqueness of $T \rightarrow T$ implies it could only be the map $1_T\text{.}$ Clearly $1_T$ satisfies $\tau_1 1_T = \tau_1$ and $\tau_2 1_T = \tau_2\text{.}$

Now suppose we have a $\mathcal{C}$-map $B_1 \xrightarrow{\beta} T\text{,}$ which we should by the definition of terminal object. In order for this map to be a map in $\mathcal{C}_{B_1 B_2}\text{,}$ we’d need a pair of maps $B_1 \xrightarrow{\beta_1} B_1$ and $B_2 \xrightarrow{\beta_2} B_2$ which preserve the structure of the following diagram:

$Structure preservation of $B_1 \rightarrow T$$

These maps need to preserve the structure of the relations $\tau_1 \beta = \beta_1$ and $\tau_2 \beta = \beta_2\text{.}$ Since $\beta \circ \tau_1$ is a map from $T \rightarrow T\text{,}$ of which there is only one, we’re forced to conclude that $\beta \circ \tau_1 = 1_T\text{.}$ Precomposing our structure preserving relations with $\tau_1$ gives us:

\begin{equation*} \tau_1 \beta \tau_1 = \tau_1 1_T = \tau_1 = \beta_1 \tau_1 \end{equation*}

\begin{equation*} \tau_2 \beta \tau_1 = \tau_2 1_T = \tau_2 = \beta_2 \tau_1 \end{equation*}

If we take $\tau_2 = \beta_2 \tau_1$ and precompose with $\beta$ again, we get $\tau_2 \beta = \beta_2 \tau_1 \beta$ which simplifies to $\beta_2 = \beta_2 \beta_1$ whichs implies that $\beta_1 = 1_{B_1} = \tau_1 \circ \beta\text{.}$ It would also follow that $\beta_2$ is uniquely determined by the sequence of maps $B_1 \xrightarrow{\beta} T \xrightarrow{\tau_2} B_2\text{.}$ Essentially, since the objects $T,B_1,B_2$ are all isomorphic to each other in $\mathcal{C}$ we can use the uniqueness of the maps there to guarantee the uniqueness of our maps in $\mathcal{C}_{B_1 B_2}\text{.}$

I’m thinking that the idea here is that there is not just a bijection between the objects $X$ and $T\text{,}$ but one between the pairs of maps $\langle \phi_1, \phi_2 \rangle$ and $\langle \tau_1, \tau_2 \rangle$ themselves. Consequently, any bijection between $\boxed{P,\pi_1,\pi_2} \leftrightarrows \boxed{T,\tau_1,\tau_2}$ automatically defines an unique bijection $\boxed{X,\phi_1,\phi_2} \leftrightarrows \boxed{P,\pi_1,\pi_2}$ through composition.

Example 5.18.2. Exercise 2: (part 1/?).

Try to create the definition of a ’sum’...

Solution.

I attempted something similar to this back in Article 4 Exercise 19. We let $I$ be a “possibly empty” set of indices and let $C_i$ be a “possibly duplicated” object in the category. This gave us the following definition of “sum family”: Definition 5.7.4

I’m not sure if I need all that for this exercise though. I think I might get away with using only the definitions of initial object, terminal object, and sum. I’m going to mirror the wording of product from Session 21.

Definition 5.18.3.

$S$ is an initial object of $\mathcal{C}$ if for every object $X$ of $\mathcal{C}$ there is exactly one $\mathcal{C}$-map $S \rightarrow X\text{.}$

Definition 5.18.4.

$T$ is an terminal object of $\mathcal{C}$ if for every object $X$ of $\mathcal{C}$ there is exactly one $\mathcal{C}$-map $X \rightarrow T\text{.}$

Definition 5.18.5.

Suppose that $A$ and $B$ are objects in a category $\mathcal{C}\text{.}$ A sum of $A$ and $B$ (in $\mathcal{C}$) is:

an object $S$ in $\mathcal{C}$
a pair of maps, $A \xrightarrow{s_1} S,B \xrightarrow{s_2} S$ in $\mathcal{C}$ satisfying: for every object $I$ and every pair of maps $A \xrightarrow{r_1} I, B \xrightarrow{r_2} I\text{,}$ there is exactly one map $S \xrightarrow{r} I$

Pictorially:

$Commutative diagram of sum$

I think we’ll also need the following uniqueness theorem, which we proved in Article 4 Exercise 7:

Theorem 5.18.6.

If $S,I$ are both initial in $\mathcal{C}$ then the unique map $S \rightarrow I$ is an isomorphism.

In the category $\mathcal{S}\text{,}$ we can think of our initial object $I$ as the empty set $\emptyset\text{.}$ There is a unique map from $\emptyset$ to any other object in $\mathcal{S}$ because there’s no elements for any arrows to originate at. Likewise, we can think of the terminal object in $\mathcal{S}$ as the singleton $\mathbf{1}$ since there’s only one object for the arrows to point to.

By substituting $A = \emptyset$ and $B = \mathbf{1}\text{,}$ we can form the following diagram:

$Sum of initial and terminal object$

Since $i_1$ is a unique map, it follows that $i_1 = i_2 \circ i_0\text{.}$ Likewise, we must also have $j_1 = j \circ i_1$ and $j_2 = j \circ i_2\text{.}$ We can subsitute our expressions to get

\begin{equation*} j_1 = j \circ (i_2 \circ i_0) = (j \circ i_2) \circ i_0 \end{equation*}

\begin{equation*} \implies j_1 = j_2 i_0 \end{equation*}

There is no map from $\mathbf{1} \rightarrow \emptyset$ in $\mathcal{S}\text{,}$ so the only possible choice of $I$ for the diagram above is the sum $\emptyset+\mathbf{1}$ with $j = 1_{\emptyset+\mathbf{1}}$ respectively.

There’s a lot to unpack here but I’m wondering if the importance is that looping the sequence $S \rightarrow I \rightarrow S \rightarrow I$ provides a way to construct the monoid $\mathbb{N}\text{.}$ Given a map $\mathbb{N} \rightarrow I\text{,}$ the compositions $\mathbb{N} \rightarrow I \rightarrow C_i$ would give us all the information we need about the structure of $X\text{.}$

I think that’s a good place to stop for the week and let this sink in. Until next time!