L&S Cat Theory: Attempted solutions to "Conceptual Mathematics" by Lawvere and Schanuel

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 $\beta (y_1)=\beta (y_2)\text{.}$

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 $f\circ \alpha =\beta \circ f\text{.}$

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 ${\beta }^5(y_1)=f({\alpha }^5{x}_1)=f({\alpha }^5{x}_2)={\beta }^5(y_2)\text{.}$

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’.

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

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 $X\text{.}$

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

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:

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, ${\alpha }^2{x}_1=x_3\ne x_1$ implies that ${\alpha }^2\ne 1_X\text{.}$

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".