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

Suppose that \(S,j_1,j_2\) and \(\bar{S},\bar{j_1},\bar{j_2}\) are both sums of \(B_1,B_2\) in a given category.

The definition of \(S\) as a sum says that for any two maps \(B_1 \xrightarrow{f_1} X\) and \(B_2 \xrightarrow{f_2} X\text{,}\) there is exactly one map \(S \xrightarrow{f} X\) such that \(f_1 = f j_1\) and \(f_2 = f j_2\text{.}\) This shows that the following diagram commutes:

By choosing \(X = \bar{S}\text{,}\) there must be a unique map \(S \xrightarrow{h} \bar{S}\) such that \(\bar{j_1} = h j_1\) and \(\bar{j_2} = h j_2\text{.}\)

We’d like to demonstrate that this map is an isomorphism, so we need to to find an inverse to it. Since \(\bar{S}\) is also a product in the category, this defines a unique map \(\bar{S} \xrightarrow{k} S\) for which \(j_1 = k \bar{j_1}\) and \(j_2 = k \bar{j_2}\text{.}\)

Let’s make the claim that \(h,k\) are inverses. Namely, that \(k h = 1_S\) and \(h k = 1_\bar{S}\text{.}\) Consider the following compositions:

The map \(k h\) composed with \(j_1\) gives \(j_1\) and composed with \(j_2\) gives \(j_2\text{.}\) By choosing \(f_1 = j_1\) and \(f_2 = j_2\text{,}\) the definition of sum says that there’s a unique map satisfying \(f j_1 = j_1\) and \(f j_2 = j_2\text{.}\) Since this is uniquely satisfied by \(f = 1_S\text{,}\) we’ve can conclude that \(k h = 1_S\text{.}\)

Likewise, we can reverse the roles of \(S,\bar{S}\) and examine the following compositions:

By the same reasoning, the unique map \(g\) for which all \(g \bar{j_1} = \bar{j_1}\) and \(g \bar{j_2} = \bar{j_2}\) could only be the unique identity map \(g = 1_\bar{S}\text{.}\) It follows that \(h k = 1_\bar{S}\) and completes the proof that \(S\) is isomorphic to \(\bar{S}\text{.}\)

Having proved the result, I still feel like there’s a weak spot in my understanding of that last step. I’m thinking it has to do with the properties of “epimorphisms” and “monomorphisms” in relation to the identity map. In the proof for the product, we rely on the fact that pre-composing by \(hk\) didn’t change anything while in the proof for sum we rely on the fact that post-composing by \(k h\) didn’t change anything. My hunch is that this is related to the connection between the section, retraction, and inverse. If an inverse map exists, then the section and retraction are both equal to it.

Part (a):

Let’s draw the left and right side separately, then compare results. Using the depiction of the sum, we can draw \(D+D\) as the following map:

Next, we can draw out \(2 \times D\text{:}\)

Having these two diagrams, how do I then compare them for equality? I think the idea is to show they can be factored through the terminal object. In particular, the maps from arrows to the terminal object form a bijection between the arrows in the sum and product.

The important characteristic I notice in this diagram is that that while we have 2 maps \(A \rightarrow T\) on each side, we have 4 maps \(D \rightarrow T\) and on the left and 5 maps \(D \rightarrow T\) on the right. This makes me wonder if two of the dots in the product need to collapse down to a single dot somehow. I’m wondering if I can resolve this difference by treating the terminal object as one of the elements in \(2\text{.}\)

I’m also not yet accounting for the “self-loops” either. I’m thinking the way to resolve this is to treat the self-loops in the products as the identity maps on \(B_1,B_2\text{.}\) I can group the two sets \(B_1,B_2\) on the left into a the sum \(B_1+B_2\) and form a bijection with my product \(2 \times D\text{.}\)

Putting both of these changes together gives us the following diagram:

Now every “dot” maps to a “self-loop”. Since there’s exactly one map \(\boxed{\bullet} \rightarrow \boxed{\bullet \circlearrowleft}\) our isomorphism above must be the “only” one. I say “only” in quotes, because the chose of assigning \(p_1,p_2\) to \(j_1,j_2\) is arbitrarily chosen but any choice is isomporphic to the other. There’s a unique map \(\alpha\) such that \(\alpha \langle p_1, p_2 \rangle = \langle p_2, p_1 \rangle\) and this map would have corresponding \(\beta\) such that \(\beta \langle j_1, j_2 \rangle = \langle j_2, j_1 \rangle\text{.}\)

Now let’s look at Part (b), which says \(D \times D = D\text{.}\) It seems like it should be obvious from the diagram of the graph product:

Every shape in the product is uniquely determined based on the point that the product is mapped to. Our map is fully determined by the domain and codomain.

For Part (C), we’ve already seen the graph of \(D+D\text{,}\) so that just leaves \(A \times D\) which can be depicted as follows:

I think in this case we establish the bijection between \(j_1,j_2\) and the inherent source and target maps of the arrow \(s,t\text{.}\) However, it’s not exactly clear to me how quite yet so lets try to factor through \(\mathbf{1}\) again.

Like before, maybe the key here is to group the terminal object and dot together as a single object. This would give us the following isomorphism:

And that complete our proof! Or at least, I think it does...