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Stream: learning: reading & references

Topic: Harold Simmons, “An Introduction to Category Theory”

Julius (Aug 16 2023 at 11:35):

I am focusing on this book as my preferred introductory-level category theory book.

Exercise 1.1.2 is about showing that a monoid is a category. I understand the general concept, but I realized I was unsure about an aspect of the identity.

I am not currently sure if every arrow in a category could have two identity arrows (a left and right), or if every identity arrow must be two-sided.

Similarly, while I assume a monoid has a unique, two-sided identity element, I wonder if there is a structure where each element has a left and right identity.

I was browsing on Wikipedia and there’s a chance this has something to do with the Grothendieck group.

Is it possible to relax the condition that a monoid have a two-sided identity; and if so, what kind of structure is that?A9046F42-5C36-4A00-9490-48DE1F26F774.png

Reid Barton (Aug 16 2023 at 12:01):

You can use the same kind of argument you used here: https://categorytheory.zulipchat.com/#narrow/stream/229199-learning.3A-questions/topic/Riehl.2C.20.E2.80.9CCategory.20Theory.20in.20Context.E2.80.9D.2E/near/382361308

Julius (Aug 16 2023 at 15:43):

I was just thinking about that, really interesting to connect the two. I’ll think more about that, thank you.

Julius (Aug 16 2023 at 15:47):

I could use help with this exercise. 48D62942-1CDA-438E-AE89-BA9DFF4E342F.png

I am able to show that sets and functions form a category. But I am not sure how to handle the requirement that f \circ \alpha = \beta circ \f.

Is there some sort of insight about these functions \alpha and \beta I need to have?

Ralph Sarkis (Aug 16 2023 at 15:54):

First of all, you need to guess what should be the identities and the composition of morphisms in this category. Do you see what they should be?

For intuition, this example of category looks a lot like the category of groups. A group is a set equipped with a binary operation (multiplication), a unary operation (an inverse) and a constant (the netural element), and group homomorphisms are functions that preserve this structure. The objects of $\mathbf{Pno}$ are sets equipped with a unary operatin and a constant, and the morphisms are function that preserve this structure.

Julius (Aug 16 2023 at 16:08):

The identities seem simple enough. Each object is a triple (A, alpha, a). The arrows are functions from A to B which preserve the nominated element and also the law f of alpha = beta of f. So it could be written id: (A, alpha, a) -> (A, alpha, a). Clearly, id(a) = a. Perhaps alpha must be the identity function, in this case?

Alpha is like an endomorphism, it’s a mapping from the set to itself. You compare it to how each element in a group can be mapped to its inverse, I think. But in b), they point out how succ: n -> n+1 could also be an example of such an alpha function.

So I guess there is some property of alpha I need to realize, to be able to argue that if f and g have the property that f of alpha = beta of f, then it would still hold for the composition of g and f.

Julius (Aug 16 2023 at 16:09):

f of alpha = beta of f
g of beta = gamma of g

And somehow show that

h of alpha = gamma of h

Julius (Aug 16 2023 at 16:15):

I think I might have it, let me just experiment with some algebraic manipulation. Thx.

Ralph Sarkis (Aug 16 2023 at 16:16):

What is the identity group homomorphism? Use this idea to define the identity for these objects.

$\alpha$ can really be any function $A \to A$, you cannot restrict it because your definition of identity morphism should work for any object $(A,\alpha,a)$. Given such an object, you should come up with the definition of $\mathrm{id}_{(A,\alpha,a)}$.

Your intuition about composition seems a bit better.

Julius (Aug 16 2023 at 18:00):

So the only way you could respect the law f of alpha = alpha of f is if f is the identity function.

Ralph Sarkis (Aug 16 2023 at 18:09):

No, the identify function is one such function that satisfies $f \circ \alpha = \alpha \circ f$ (and $f(a) = a$), but there are others. Still, we have to make sure the identity function satisfies it, so we can define the identity morphism $(A,\alpha,a) \to (A,\alpha, a)$ to be the identity function. (If the identity function did not satisfy this condition, we would have to look for another candidate to be the identity morphism.)

Now, for composition, if we have two morphisms $f: (A,\alpha, a) \to (B,\beta, b)$ and $g: (B,\beta, b) \to (C, \gamma, c)$, we have to define their composite. Since the identity morphism was the identity function, we may try to define the composition just like the composition of functions, i.e. $(g \circ f)(x) = g(f(x))$ for all $x \in A$. Can you show that this definition respects the condition to be a morphism $(A,\alpha, a) \to (C, \gamma, c)$.

Julius (Aug 16 2023 at 20:17):

Thanks. I feel like I understand the algebra but I would like a more concrete understanding of the function composition equation. I feel like I haven’t encountered some real-world functions which obey this equation. I wrote my thoughts on this problem in this PDF. I’d love if someone could take a look. Thank you.Pno_category.pdf

Ralph Sarkis (Aug 16 2023 at 20:25):

If you forget about the binary operation of a group, you get an object of Pno, the alpha is the inverse function and the a is the identity element of the group. So a function is a morphism when it preserves "taking inverses" (it does not have to preserve the multiplication if you only care about it being a morphism in Pno).

Another intuitive way to understand objects of Pno is as dynamical systems. A is the set of states that the system can be in, alpha is takes on state and gives back the next state after one step of evolution of the system, and a could be the initial state of the system.

A morphism between two such systems is a function between the states that maps the initial state in A to the initial state in B and for every state, applying the map then taking a step is the same thing as taking a step then applying the map.

David Egolf (Aug 16 2023 at 20:38):

Here's a related perspective, based around the idea that morphisms between objects should transform true equations into analogous true equations.

Imagine we have the true equation that $\alpha(r) = s$ in the context of $(A, \alpha, a)$, where $r,s \in A$. In general, we like morphisms between structures to send true equations to analogous true equations. So, we'd like to be able to change each part of this true equation to get a true equation in the context of $(B, \beta, b)$. That is, given a function $f: A \to B$, we'd like $\beta(f(r)) = f(s)$ to be true provided that $\alpha(r)=s$ is true. Here I've mapped each element of $A$ over to a corresponding element of $B$ using $f$, and I've changed $\alpha$ out for the corresponding function $\beta$ that belongs to $(B, \beta, b)$.

Let's find a condition we can place on $f$ to get our desired behaviour described above. Let us assume that $\alpha(r) = s$ is true. Applying $f$ to each side, we find that $f(\alpha(r)) = f(s)$. Now, we want to show that $\beta(f(r)) = f(s)$ must also be true. If $f \circ \alpha = \beta \circ f$, then $f(s) = f(\alpha(r)) = \beta(f(r))$, which is what we want to be true!

So, the condition $f \circ \alpha = \beta \circ f$ ensures that true equations involving $\alpha$ in $(A, \alpha, a)$ get mapped over by $f$ to corresponding true equations involving $\beta$ in $(B, \beta, b)$.

Jencel Panic (Aug 20 2023 at 06:36):

image.jpg

Jencel Panic (Aug 20 2023 at 06:53):

@Ralph Sarkis already explained everything, but just to reiterate:

To prove that morphisms compose, you need three object in your category (A, B, C) and two morphisms ($f$ and $g$). And you need to give is a way to make a third morphism from those two, such that applying it is equivalent to applying the two morphisms ($f$ and $g$) one after the other.

In this case, the way to make the third morphism is obvious, since $f$ and $g$ are just normal functions between sets (normal functional composition).

What is not so obvious at first is whether applying this third moprhism ($g \circ f$) is equivalent to applying $f$ and $g$. Or actually scratch that, what is not obvious is whether $g \circ f$ is a valid morphism in Pno i.e. whether $g \circ f$ would preserve the extra law that is needed for our structures to be a category i.e. whether $\gamma \circ( g \circ f) =( g \circ f) \circ \alpha$.

The diagram shows that both of these things are true.

Jencel Panic (Aug 20 2023 at 14:57):

image.jpg

Jencel Panic (Aug 20 2023 at 15:04):

Also, I think it's apt to point out that these commuting squares are actually natural transformations. If we take $\mathbb{N}$ to be a category with one object and one endomorphism, then any pair containing a set and an endofunction, can be represented by a functor $\mathbb{N} \to Set$. Under this representation, $f$ and $g$ are natural transformations.