Nice work in [comment #5](https://forum.azimuthproject.org/discussion/comment/18341/#Comment_18341), 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](https://forum.azimuthproject.org/discussion/comment/18341/#Comment_18341) 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](https://forum.azimuthproject.org/discussion/comment/18346/#Comment_18346). 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

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

by

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

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.