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

Our basic idea here is that \(f\) "respects the structure" of our category \(\mathcal{S}^{\circlearrowright}\text{.}\) That is:

Satisifies \(\boxed{f \circ \alpha = \beta \circ f}\) for all endomaps \(\alpha\) and \(\beta\text{.}\)

Now suppose we have another map \(g\) with

We can then compose these maps to produce a composition \(g \circ f\text{:}\)

This map should preserve structure also if we can prove that \((g \circ f) \circ \alpha = \gamma (g \circ f)\text{.}\) This should be fairly straight forward to show using the associative property and our previous assumptions:

And that completes the exercise.

Recall that an endomap \(e: A \longrightarrow A\) is "idempotent" if \(e \circ e = e\text{.}\) If \(r\) is a "retraction" for \(e\text{,}\) that implies \(r \circ e = 1_A\text{.}\)

Watch what happens when if we apply \(e\) on the right of both sides to \(r \circ e = 1_A\text{:}\)

Essentially, a retraction preserves the fixed point proprerty. Since \(e\) fixes every point it touches and this retraction can restore each value afterwards, that can only happen if \(e = 1_A\text{.}\)

I’m not really sure what the question is, so I’m assuming this is a "prove it" problem. I think this is essentially the approach I tried to take with Brouwer’s theorems lasts.

If we define as set \(A\) being "even" iff there is an involution with no fixed points. This is like the "antipodal map" I explored earlier that pairs up each point in a space with it’s "opposite". When we’re working with an antipodal map on the interval, we either we have a point in the middle that maps to itself or we "jump" over it. The fixed point property becomes a proxy for whether or not the map is continuous because we forced the interval to be "odd". The only problem with using this idea here is that the interval had a natural order to it while the set \(A\) may not, so our involution isn’t necessarily unique.

Maybe it might help to think of taking a map \(f\) applying and taking an "image" of it for every \(\mathbf{1} \xrightarrow{x} A \xrightarrow{f} \text{Image}_f(A)\text{.}\) Either we find some point \(f x = x\) or we don’t, but we can use this to classify each point as "fixed" or "not". Essentially this process defines map \(g: A \longrightarrow\{0,1\}\) that returns \(1\) if a point is fixed by \(f\) or \(0\) if it doesn’t. However, since \(f\) is known to be an involution with \(f \circ f = 1_A\) then there needs to be at least as many points in \(\text{Image}_f(A)\) as there are in \(A\) which ensures that \(\text{Image}_f(A) = A\text{.}\)

Framing the problem like this allows us to ask if there exists a section of \(A\) which permits a retraction on \(g\text{.}\) More specifically, we’re looking for a map \(s: \{0,1\} \longrightarrow A\) with \(g \circ s = 1_{\{0,1\}}\text{.}\) If \(f = 1_A\) then \(\forall x \in A: g x = 1 \text{,}\) and consequently \(s(0)\) would be undefined. If \(f\) has no fixed points then \(\forall x \in A: g x = 0\text{,}\) and consequently \(s(1)\) would be undefined. In order for this map \(s\) to exist, \(f\) must have at least one fixed point and one non-fixed point. That is, for some \(s(1) = x \in A\) we have \(f x = x \) and some \(s(0) = y \in A\) such that \(f y \neq y\) with \(g x = 1\) and \(g y = 0\text{.}\)

The problem with this is that such a map can’t exist without \(A\) having at least three points: \(x\text{,}\) \(y\text{,}\) \(f y\text{.}\) It’s impossible for a composition of the any form \(\{x,y,fy\} \longrightarrow \{0,1\} \longrightarrow \{x,y,fy\}\) to ever produce an identity map. If a retraction for \(g\) existed it would need to be the same as the section, so clearly the fixed point property on \(f\) means that no map is a retraction from \(A\) to \(\{0,1\}\text{.}\)

I feel like I’m starting to talk in circles now, so I’ll just add that I think there’s some parallels to be made with the "split idempotent" exercise from Session 9.

If all points in \(f\) are fixed, that implies a splitting of \(f\) by \(\mathbf{1}\text{.}\) If no points in \(f\) are fixed, that also implies a splitting of \(f\) by \(\mathbf{1}\text{.}\) If \(f\) has both fixed and non-fixed points, that implies a splitting of \(f\) by \(\{0,1\}\text{.}\) Since there is no isomorphism between \(\mathbf{1}\) and \(\{0,1\}\text{,}\) we use the result of Session 9 Exercise 3 to infer that \(f\) can’t be split by both of them simultaneously. If our \(f\) has both fixed and non-fixed points, the set of fixed points would preserve an isomorphism with \(\mathbf{1}\) — meaning that the point fixed by \(f\) would necessarily be unique. Likewise, each non-fixed point \(y\) needs to exists as part of a pair with some \(f y\) to preserve the isomorphism between \(\{0,1\}\) and \(\{y,fy\}\text{.}\)

I’m not sure if I’m using this concept of "splitting \(f\) by \(\{0,1\}\)" correctly here, but it seems like it might be a reasonable analogy of "division by two" for generic maps.

This \(\alpha\) would be characterized as an "involution" since \(-(-x)=x\) implies \(\alpha \circ \alpha = 1_\mathbb{Z}\text{.}\) It has a fixed point at \(0\) since \(-0 = 0\text{.}\) It’s not idempotent since \(\alpha \circ \alpha \neq \alpha\) for any \(x \lt 0\text{.}\)

This \(\alpha\) would be characterized as an "idempotent" since \(||x|| = |x|\) implies \(\alpha \circ \alpha = \alpha\text{.}\) It has an infinite number of fixed points, since \(\forall x \in \mathbb{Z}\) such that \(x \geq 0\) we have \(\alpha(x) = x\text{.}\) It is not an involution since \(\alpha \circ \alpha(x) \neq x\) for any \(x \lt 0\text{.}\)

The previous page defines "automorphism" as an "endomap which is also an isomorphism", so I’m going to have to say "yes". We can define the inverse by \(\alpha^{-1}(x) = x - 3\text{.}\) It follows that \(\alpha \circ \alpha^{-1}(x) = \alpha(x - 3) = (x - 3) + 3 = x\) and \(\alpha^{-1} \circ \alpha(x) = \alpha^{-1}(x + 3) = (x + 3) - 3 = x\text{,}\) satisfying the definition for an isomorphism. This map is neither idempotent nor an involution, and has no fixed points.

Conceptually, we can loosely define an inverse by \(\alpha^{-1}(x) = \lfloor x/5 \rfloor\text{.}\) However, this will not meet the definition for an automorphism. It mights the criteria for a "section" since \(\alpha^{-1}(x) \circ \alpha = \lfloor (5x)/5 \rfloor = x\text{,}\) but it’s not a "retraction" since \(\alpha \circ \alpha^{-1}(1) = 5 \lfloor x/5 \rfloor \neq x\) for any \(x\) which is not a multiple of 5. This map is neither idempotent nor an involution, but does have fixed point at 0.

We’re given an internal diagram for \(\alpha\text{,}\) and I took the liberty of naming our three points \(\{x,y,z\}\) as follows:

Clearly the above diagram of \(\alpha\) is fully described by the relations \(\alpha(x)=y\text{,}\) \(\alpha(y)=z\text{,}\) and \(\alpha(z)=y\text{.}\)

In order to demonstrate that \(\alpha^3=\alpha\text{,}\) we just need to show it holds for all three points: that \(\alpha^3(x) = \alpha(x)\text{,}\) \(\alpha^3(y) = \alpha(y)\text{,}\) and \(\alpha^3(z) = \alpha(z)\text{.}\) Let’s verify these one at a time:

Having established that \(\alpha^3=\alpha\) holds for all three values, let’s next address idempotency by checking if \(\alpha^2=\alpha\text{.}\) It’s pretty clear that this fails for \(x\text{.}\) \(\alpha^2(x) = \alpha(y) = z\) but \(\alpha(x) = y\text{,}\) so \(\alpha^2(x) \neq \alpha(x)\text{.}\)

Finally, we need to check whether or not \(\alpha\) is an involution. Since an involution would imply \(\alpha \circ \alpha = 1_A\text{,}\) the case of \(\alpha^2(x) = z \neq x\) serves as a counter-example to this expression.