Article 2: Isomorphisms, Part 2

Section 3.2 Article 2: Isomorphisms, Part 2

Choosing to break this article into subsections means that I can take my time with the only excercise. With that in mind, I’m going to use the opportunity to work on improving my LaTeX skills by attempting to enumerate the diagrams of these solutions instead of just counting them. Sounds like fun, huh?

Example 3.2.1. Exercise 5.

...how many maps \(f\) are there with \(g \circ f = 1_{\set{0,1}}\text{?...}\)

Solution.

Let’s begin by reproducing the internal diagram of \(g\text{:}\)

Figure 3.2.2. Recreation of diagram from Article II, Exercise 5

Since I want \(g \circ f = 1_{\set{0,1}}\text{,}\) I know that \(f\) needs to send each element of \(\set{0,1}\) to something that will return to that value after applying \(g\text{.}\) It seems reasonable to assume that the "possible arrows" for \(f\) must come from the reversed arrows of \(g\) as shown here:

Figure 3.2.3. Diagram from Article II, Exercise 5 but with reversed arrows

Since there are three options for where \(f\) sends \(0\text{,}\) two options for where it sends \(1\text{,}\) and these choices are independent of each other, the total number of possibilities is the product: \(3 \times 2 = 6\text{.}\) This lends them nicely to a 3x2 grid where the column is the choice of where to send \(0\) and row is the choice of where to send \(1\text{.}\)

Figure 3.2.4. Grid of potential solutions for \(f\) in Article II, Exercise 5

Now, that we’ve drawn all possible choices for \(f\text{,}\) let’s choose one and see how many options we might have for \(g\) with the same property \(g \circ f = 1_{\set{0,1}}\text{.}\) I’ll start by taking a single one:

Figure 3.2.5. One possible solution to \(f\) from Article II, Exercise 5

Like before, I know my map \(g\) must contain the reverse of the arrows from my chosen map \(f\text{,}\) but fixing those arrows still leaves 3 points that are unaccounted for in the map. For each of those 3 possible sources, there are two possible destinations.

Figure 3.2.6. Possible choices for \(g\) given a choice of \(f\)

Since those choices are independent of each other, that means the total number of combinations is \(2 \times 2 \times 2 = 2^3 = 8\text{.}\)

Figure 3.2.7. Grid of potential maps for \(g\) given my chosen \(f\)

With that, we’ve completed the exercise (and then some).