Article 2: Isomorphisms, Part 3

Section 3.3 Article 2: Isomorphisms, Part 3

As I wrap up Article 2, I keep thinking back to the last expression in Exercise 4 and wonder if there’s a better way of articulating what I saw. I’m starting to see more of the connection with the injective (one-to-one) and surjective (onto) properties I typically associate with isomorphisms are effectively guaranteed by the definitions of retraction and section.

I also find the symbol choice of \(T\) in this part of Article 2 somewhat curious. The programmer in me wants to associate \(T\) with a "type" or "template", while the mathematician in me is quietly screaming "TOPOLOGY!". I suppose I’ll have to keep going to find out who’s right-er.

Example 3.3.1. Exercise 6.

If the map \(A \xrightarrow{f} B\) has a retraction, ...

Solution.

Essentially the argument is the same as the preceeding proof with minor substitutions. We’re given that \(A \xrightarrow{f} B\) has a retraction, so we have some \(r\) with the property \(r \circ f = 1_A\text{.}\)

Figure 3.3.2. Possible compositions given \(f\) has a retraction.

This external diagram implies that we can define a map \(B \xrightarrow{t} T\) by using the composition \(t = g \circ r\text{.}\) All that remains is to verify that it satistfies the required equation:

\begin{equation*} t \circ f = g \circ r \circ f = g \circ 1_A = g \end{equation*}

And that completes the proof.

Example 3.3.3. Exercise 7.

Suppose the map \(A \xrightarrow{f} B\) has a section, ...

Solution.

Let’s get our bearings straight with a diagram. We know \(f\) has a section \(s\) satisfying \(f \circ s = 1_B\text{.}\) We also have a set \(T\) with a pair of maps \(B \xrightarrow{t_1} T\) and \(B \xrightarrow{t_2} T\text{.}\)

Figure 3.3.4. External diagram of given information in Article 2 Exercise 7.

Now lets further assume that \(t_1 \neq t_2\text{,}\) so that we have some \(b \in B\) such that \(t_1(b) \neq t_2(b)\text{.}\) Consider the point defined by \(s(b)=x\text{.}\) If \(t_1 \circ f = t_2 \circ f\text{,}\) then it must also hold true that \((t_1 \circ f)(x) = (t_2 \circ f)(x)\) for all \(x\text{.}\) If we expand this, we get the following:

\begin{equation*} (t_1 \circ f)(x) = (t_2 \circ f)(x) \end{equation*}

\begin{equation*} (t_1 \circ f)(s(b)) = (t_2 \circ f)(s(b)) \end{equation*}

\begin{equation*} (t_1 \circ f \circ s)(b) = (t_2 \circ f \circ s)(b) \end{equation*}

\begin{equation*} (t_1 \circ 1_B)(b) = (t_2 \circ 1_B)(b) \end{equation*}

\begin{equation*} t_1(b) = t_2(b) \end{equation*}

This contradicts our earlier assumption that \(t_1 \neq t_2\) and completes the proof. It’s a little easier to see what’s going on if we rearrange the diagram some:

Figure 3.3.5. Diagram of compositions induced by assumptions of Article 2 Exercise 7.

As long as \(t_1 \circ f = t_2 \circ f\text{,}\) then \(t_1 \circ f \circ s = t_2 \circ f \circ s\text{.}\) Since we’ve defined \(s\) as a section of \(f\text{,}\) it means that \(f \circ s = 1_B\text{.}\) Therefore, \(t_1 \circ f \circ s = t_2 \circ f \circ s\) holds if and only if \(t_1 = t_2\text{.}\)

Example 3.3.6. Exercise 8.

...the composite of two maps, each having sections, ...

Solution.

Suppose we have composable maps \(A \xrightarrow{f_1} B\) and \(B \xrightarrow{f_2} C\) with respective sections \(B \xrightarrow{s_1} A\) and \(C \xrightarrow{s_2} B\text{.}\) By definition, \(f_1 \circ s_1 = 1_B\) and \(f_2 \circ s_2 = 1_C\text{.}\)

Based on this information, the composition \(A \xrightarrow{f_2 \circ f_1} C\) has a section determined by the reversed composition of the sections \(C \xrightarrow{s_1 \circ s_2} A\text{.}\) All we need to do is to show that \((f_2 \circ f_1) \circ (s_1 \circ s_2) = 1_C\text{.}\)

\begin{equation*} (f_2 \circ f_1) \circ (s_1 \circ s_2) \end{equation*}

\begin{equation*} = f_2 \circ (f_1 \circ s_1) \circ s_2 \end{equation*}

\begin{equation*} = f_2 \circ 1_B \circ s_2 \end{equation*}

\begin{equation*} = f_2 \circ s_2 \end{equation*}

\begin{equation*} = 1_C \end{equation*}

It’s easy to see why this order reversal is necessary by examining a diagram:

Figure 3.3.7. External diagram of given information in Article 2 Exercise 8.

Example 3.3.8. Exercise 9.

Suppose \(r\) is a retraction of \(f\text{...}\)

Solution.

Knowing that \(r\) is a retraction of \(f\) tells us that \(r \circ f = 1_A\text{.}\) If we define the map \(e = f \circ r\text{,}\) we can prove that \(e \circ e = e\) through the use of the associative property.

\begin{equation*} e \circ e = (f \circ r) \circ (f \circ r) \end{equation*}

\begin{equation*} = f \circ (r \circ f) \circ r \end{equation*}

\begin{equation*} = f \circ 1_A \circ r \end{equation*}

\begin{equation*} = f \circ r = e \end{equation*}

That completes the proof that \(e\) is idempotent.

If we further assume that \(f\) is an isomorphism, then we have some inverse function \(f^{-1}\) such that \(f^{-1} \circ f = 1_A\) and \(f \circ f^{-1} = 1_B\text{.}\) Since this inverse is unique, it follows that \(r = f^{-1}\text{.}\) Consequently, \(e = f \circ r = f \circ f^{-1} = 1_B\text{.}\)

Example 3.3.9. Exercise 10.

If \(A \xrightarrow{f} B \xrightarrow{g} C\) are both isomorphisms...

Solution.

I think this one is a little easier to see with a diagram.

Figure 3.3.10. External diagram of given information in Article 2 Exercise 10.

Basically, to go backward to \(C\) from \(A\text{,}\) the order of inverses needs to also be reversed. We can prove this algebraically also.

Since \(f\) is an isomorphism, it follows that \(f^{-1} \circ f = 1_A\) and \(f \circ f^{-1} = 1_B\text{.}\) Likewise, \(g\) being an isomorphism implies \(g^{-1} \circ g = 1_B\) and \(g \circ g^{-1} = 1_C\text{.}\) The composition \(f^{-1} \circ g^{-1}\) can be shown to sasify the conditions for being the inverse of \(g \circ f\text{.}\)

First, we must demonstrate that \((f^{-1} \circ g^{-1}) \circ (g \circ f) = 1_A\text{:}\)

\begin{equation*} (f^{-1} \circ g^{-1}) \circ (g \circ f)=f^{-1}) \circ (g^{-1} \circ g) \circ f \end{equation*}

\begin{equation*} = f^{-1} \circ 1_B \circ f \end{equation*}

\begin{equation*} = f^{-1} \circ f = 1_A \end{equation*}

Second, we must demonstrate that \((g \circ f) \circ (f^{-1} \circ g^{-1}) = 1_C\text{:}\)

\begin{equation*} (g \circ f) \circ (f^{-1} \circ g^{-1}) = g \circ (f \circ f^{-1}) \circ g^{-1} \end{equation*}

\begin{equation*} = g \circ 1_B \circ g^{-1} \end{equation*}

\begin{equation*} = g \circ g^{-1} = 1_C \end{equation*}

This complete our proof that \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\)

Example 3.3.11. Exercise 11.

If \(A = \set{\text{Fatima},\text{Omer},\text{Alysia}}\text{...}\)

Solution.

Any \(f\) that maps each element in \(A\) to a unique element in \(B\) should do the trick here.

Figure 3.3.12. Hypothetical \(f\) for Article 2 Exercise 11

This \(f\) has a unique inverse \(f^{-1}\) that reverses each of these arrows:

Figure 3.3.13. Corresponding \(f^{-1}\) for Article 2 Exercise 11

If we have \(C = \set{true,false}\text{,}\) we’d be unable to define an isomorphism from \(A \longrightarrow C\) because we don’t have enough points to match them up one by one. One of those points in \(C\) will have two arrows leading in and cause us to lose information about where we came from in the process:

Figure 3.3.14. Illustration of invertability problem with \(A \longrightarrow C\)

Example 3.3.15. Exercise 12.

How many isomorphisms are there...

Solution.

Essentially, the number of isomorphisms is a permutation problem. If we decide on a fixed order of elements in \(A\text{,}\) each choice uses up the respective point in \(B\) that it gets matched to. Therefore, we have 3 choices for Fatima, 2 choices for Omer, and 1 choice for Alysia, or \(3! = 3 \times 2 \times 1 = 6\) possible choices all together.

The number of automorphisms should be the same, namely 6. It doesn’t matter if we’re assigning Fatima to \(\set{Fatima,Omer,Alysia}\) or \(\set{coffee,tea,cocoa}\text{,}\) it’s precisely the same number of choices.

The number 27 in this context is the total number of possible maps from a set of 3 elements to another set of 3 elements: \(3^3 = 27\text{.}\) This is the formula we arrived at after Article 1. Clearly "the number of invertable maps" should be less than "the total number of maps" because not every map will be invertable.

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