Session 14: Maps preserve positive properties

Section 4.9 Session 14: Maps preserve positive properties

The authors include some of these solutions in the reading, but I’m going to try to work them all out on my own. Just because.

Example 4.9.1. Exercise 1:.

Let \(x_1\) and \(x_2\) be two points of \(X\text{...}\)

Solution.

Given that \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\text{,}\) we can apply \(\beta\) on the left of both sides to get \(\beta(y_1) = (\beta \circ f)(x_1)\) and \(\beta(y_2) = (\beta \circ f)(x_2)\text{.}\) Since the structure preservation property enforces \(f \circ \alpha = \beta \circ f\text{,}\) we can subsitute \(f \circ \alpha\) into our expressions to get \(\beta(y_1) = (f \circ \alpha)(x_1) = f(\alpha x_1)\) and \(\beta(y_2) = (f \circ \alpha)(x_2) = f(\alpha x_2)\text{.}\) If \(\alpha x_1 = \alpha x_2\text{,}\) it follows directly that \(\beta(y_1) = \beta(y_2)\text{.}\)

Example 4.9.2. Exercise 2:.

If instead we know that \(x_2 = \alpha^5(x_1)\text{...}\)

Solution.

Like before, if can apply \(\beta^5\) on the left of \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\) to get \(\beta^5(y_1) = (\beta^5 \circ f)(x_1)\) and \(\beta^5(y_2) = (\beta^5 \circ f)(x_2)\text{.}\) We can expand the whole thing out and apply a repeated subsitution using \(f \circ \alpha = \beta \circ f\text{.}\)

\begin{equation*} \beta^5 \circ f = \beta \circ \beta \circ \beta \circ \beta \circ \beta \circ f \end{equation*}

\begin{equation*} = \beta \circ \beta \circ \beta \circ \beta \circ f \circ \alpha \end{equation*}

\begin{equation*} = \beta \circ \beta \circ \beta \circ f \circ \alpha \circ \alpha \end{equation*}

\begin{equation*} = \beta \circ \beta \circ f \circ \alpha \circ \alpha \circ \alpha \end{equation*}

\begin{equation*} = \beta \circ f \circ \alpha \circ \alpha \circ \alpha \circ \alpha \end{equation*}

\begin{equation*} = f \circ \alpha \circ \alpha \circ \alpha \circ \alpha \circ \alpha = f \circ \alpha^5 \end{equation*}

Thus \(\beta^5(y_1) = (\beta^5 \circ f)(x_1) = (f \circ \alpha^5)(x_1) = f(\alpha^5 x_1)\) and \(\beta^5(y_2) = (\beta^5 \circ f)(x_2) = (f \circ \alpha^5)(x_2) = f(\alpha^5 x_2)\text{.}\) If \(\alpha^5 x_1 = \alpha^5 x_2\text{,}\) then \(\beta^5(y_1) = f(\alpha^5 x_1) = f(\alpha^5 x_2) = \beta^5(y_2)\text{.}\)

Example 4.9.3. Exercise 3:.

If \(\alpha(x) = x\text{...}\)

Solution.

Knowing \(y = f(x)\text{,}\) just apply \(\beta\) on the left of both sides to get \(\beta y = \beta f(x) = (\beta f)(x) = (f \alpha)(x)= f(\alpha x) = f(x) = y\text{.}\) It follows that \(\beta y = y\) is also a ’fixed point’.

Example 4.9.4. Exercise 4:.

Give an example in which...

Solution.

Since we’re given a "hint" let’s use it. Consider \(X^{\circlearrowright \alpha} = \boxed{ x_1 \rightarrow x_2 \circlearrowright}\) and \(Y^{\circlearrowright \beta} = \boxed{ y \circlearrowright}\text{.}\) Our only possible map \(X^{\circlearrowright \alpha} \xrightarrow{f} Y^{\circlearrowright \beta}\) has the following diagram:

Figure 4.9.5. Only possible map \(X^{\circlearrowright \alpha} \xrightarrow{f} Y^{\circlearrowright \beta}\)

Here the point \(x_1\) has the desired properties. Since \(\alpha x_1 = x_2\text{,}\) \(x_1\) is not a fixed point in \(X\text{.}\) However, \(f x_1 = y\) and \(\beta y = y\) is a fixed point of \(Y\text{.}\) These fixed points are a "one-way" relationship. Since \(x_2\) is fixed in \(X\) then \(f x_2\) must be fixed in \(Y\text{,}\) but just because \(f x_1\) is fixed in \(Y\) doesn’t mean \(x_1\) is fixed in \(X\text{.}\)

Example 4.9.6. Exercise 5:.

Show that if \(\alpha^4(x) = x\text{...}\)

Solution.

Based on what we’ve seen so far, I’m thinking that these properties might be satisfied by the following map \(X^{\circlearrowright \alpha} \xrightarrow{f} Y^{\circlearrowright \beta}\text{:}\)

Figure 4.9.7. Map from a 4-cycle to a 2-cycle

Essentially, I’m taking the 4-cycle and splitting it in half. Perhaps the best way to illustrate that we get the desired properties is with a table:

Table 4.9.8. Validation table for Session 14 Exercise 5

\(\mathbf{1} \xrightarrow{x} X\)	\(\alpha x\)	\(f x\)	\(f \alpha x\)	\(\beta f x\)
\(x_1\)	\(x_2\)	\(y_1\)	\(y_2\)	\(y_2\)
\(x_2\)	\(x_3\)	\(y_2\)	\(y_1\)	\(y_1\)
\(x_3\)	\(x_4\)	\(y_1\)	\(y_2\)	\(y_2\)
\(x_4\)	\(x_1\)	\(y_2\)	\(y_1\)	\(y_1\)

Since the last columns are equal, we know this is a valid \(\mathcal{S}^{\circlearrowright}\)-map. For each point in \(X\text{,}\) \(\alpha^4(x) = x\text{.}\) For each point in \(Y\text{,}\) \(\beta^2 y = y\text{.}\) However, \(\alpha^2 x_1 = x_3 \neq x_1\) implies that \(\alpha^2 \neq 1_X\text{.}\)

I can imagine that this leads into some divisibility rules for cycles. I was only able to shrink the cycle here because 4 is evenly divisible by 2. If two cycles are relatively prime in lengths, the only cycle we’d be able to map them to (and still preserve our structure) is a "fixed point".

Prev Top Next