Nice work in [comment #5](, Jonathan! I'm glad to see someone tackle these puzzles.

Sophie's point is a good one, though:

> Does this mean that you are defining \\(\leq\\) according to the amount of resources in each complex? I thought that we were using \\(\leq\\) to represent possible reactions. Remember, \\(y \leq y'\\) means that we can get \\(y\\) from \\(y'\\) not that there are fewer resources in \\(y\\) than in \\(y'\\).

Nothing in [comment #5]( mentions how the ordering is defined in terms of the reactions we have available, so the argument must be incomplete, at least... though I think it's on the right track, and maybe Sophie fixed it in [comment #10]( I'll think about that tomorrow. For now, something simpler:

I think your argument would be exactly right if we were dealing with a collection of reactions that can destroy one item of each time. Given a set \\(S = \\{s_1, \dots, s_n\\} \\) and reactions

\[ s_i \to 0 \]

the preorder we get on \\(\mathbb{N}(S)\\) is the one where

\[ a_1 s_1 + \cdots a_n s_n \le b_1 s_2 + \cdots + b_n s_n \; \textrm{ iff } \; a_i \le b_i \textrm{ for all } 1 \le i \le n .\]

In other words, we can get from one complex to another only by destroying things.

Let's consider this case. Suppose we have any function \\(\phi : S \to T \\). We can use it to define a homomorphism

\[ f: \mathbb{N}[S] \to \mathbb{N}[T] \]


\[ f( a_1 s_1 + \cdots + a_n s_n) = a_1 s_{\phi(1)} + \cdots + a_n s_{\phi(n)} \]

The \\(f\\) in my puzzles is an example of this idea. But let's not solve those puzzles now: instead, give \\(\mathbb{N}[S] \\) and \\(\mathbb{N}[T] \\) the preorders described above.. Then \\(f\\) is a strict monoidal monotone. I believe Jonathan's argument shows \\(f\\) has no right adjoint unless \\(\phi\\) is onto.