Session 25: Labelings and products of graphs, Part 6

Section 5.29 Session 25: Labelings and products of graphs, Part 6

I’m gong to give these last two exercises a chance, then I’d really like to wrap this up so I can move on.

Example 5.29.1. Exercise 5:.

In this exercise, $B = \boxed{\bullet \rightarrow \bullet \rightarrow \bullet}$ and $C = \boxed{\bullet \leftrightarrows \bullet ~~~~ \bullet}\text{...}$

Solution.

So lets start by showing $B$ is not isomorphic to $C\text{.}$ Since there are two arrows in $B$ and two arrows in $C\text{,}$ any map $B \rightarrow C$ must map arrows to arrows. This means that there are only two such maps that preserve structure:

$Only two maps from B to C$

In both of these cases, the respective map takes 3 dots in $B$ and maps them to only 2 dots in $C\text{.}$ Since there is no retraction from $\mathbf{2}$ to $\mathbf{3}\text{,}$ the only two maps $B \rightarrow C$ are non-invertible. This proves that $B$ is not isomorphic to $C\text{.}$

Next, let’s look at the product $A \times B\text{:}$

$Product of A times B$

Similarly, let’s also draw out $A \times C\text{:}$

$Product of A times C$

In both of these cases, the product contains 2 disjoint naked arrows and 6 dots total. An isomorphism can be constructed by pairing arrows with arrows and dots with dots. One such map is shown below:

$Possible isomorphism$

There should be 4 isomorphisms to choose from, but we only needed one. QED.

Example 5.29.2. Exercise 6:.

Assuming that $X\text{,}$ $B_1$ and $B_2$ are objects of a category...

Solution.

Let’s start with the products $X \times B_1,p_{1,1},p_{2,1}$ and $X \times B_2,p_{1,2},p_{2,2}\text{.}$ For any pair of maps $X \xrightarrow{f_1} B_1$ and $X \xrightarrow{f_2} B_2\text{,}$ these products define a pair maps $X \xrightarrow{\langle 1_X, f_1 \rangle} X \times B_1$ and $X \xrightarrow{\langle 1_X, f_2 \rangle} X \times B_2$ . The definition of product says that there’s exactly one such pair of maps with that $p_{1,1} \langle 1_X, f_1 \rangle = 1_{X}\text{,}$ $p_{2,1} \langle 1_X, f_1 \rangle = f_1\text{,}$ $p_{1,2} \langle 1_X, f_2 \rangle = 1_{X}\text{,}$ and $p_{2,2} \langle 1_X, f_2 \rangle = f_2\text{.}$ These products can be diagrammed as follows:

$X times B_1$

$X times B_2$

Let’s now construct a third product $B_1 \times B_2, p_{1,3}, p_{2,3}$ to fit the following diagram:

$B_1 times B_2$

Next, let’s consider the sum of $B_1$ and $B_2\text{.}$ For any pair of maps $B_1 \xrightarrow{g_1} X$ and $B_2 \xrightarrow{g_2} X\text{,}$ our sum defines a pair of maps $B_1 \xrightarrow{j_{1,1}} B_1+B_2$ and $B_2 \xrightarrow{j_{2,1}} B_1+B_2$ such that there is a unique map $B_1+B_2 \xrightarrow{g} X$ satisfying $g j_{1,1} = g_1$ and $g j_{1,1} = g_2$

$Sum of B_1 and B_2$

We can combine the sum and and product diagrams to get the following:

$B_1 times B_2$

Since our maps $f_1,f_2$ and $g_1,g_2$ are arbitrary, we can choose them to satisfy $g j_1 \langle f_1 f_2 \rangle = 1_X\text{.}$

I think we’re also going to need a fourth product for $X \times (B_1 + B_2)\text{.}$ This should correspond to the following diagram:

$X times (B_1 + B_2)$

We’re also going to need another sum for $(X \times B_1) + (X \times B_2) \text{.}$ It should conform to the following diagram:

$Sum of X times B_1 and X times B_2$

Our definition sum says there is exactly one map $X \xrightarrow{j_2} (X \times B_1) + (X \times B_2)$ satisfying $h j_2 = 1_X\text{.}$

With that, I think we have enough maps to construct the one we’re looking for.

\begin{equation*} (X \times B_1) + (X \times B_2) \xrightarrow{h} X \xrightarrow{\langle 1_X, j \langle f_1 f_2 \rangle \rangle} X \times (B_1 + B_2) \end{equation*}

QED. Or, at least I hope so. I’m a little uncertain about the existence of this map $B_1 \times B_2 \xrightarrow{j_1} B_1+B_2\text{,}$ but it seems like a reasonable assumption.

And with that, I think I’m ready to push onward.