John, my understanding is that an empty reaction network is one in which only the trivial relation \\(x \le x\\) holds for all \\(x\\). Under that definition, my argument comes out even worse: I was working with a completely unrelated relation to begin with. We could imagine that one to be an empty reaction network except that reagents may (discretely) _evaporate_.

Thinking through my argument, it's only _accidentally_ correct for empty reaction networks. It's still true that we need minimal/maximal elements of \\(\mathbb{N}\\) to cover the forgotten reagents, since an infinity of \\(S\\)-complexes collapse onto the same \\(T\\)-complex. But the only element whose inverse image we need to worry about for computing \\(g(x)\\) is \\(x\\) itself.

(**EDIT:** Wait, I'm even doubting this now -- because no two of the collapsing infinity of \\(S\\)-complexes are related, we don't have a unique join.)

Also, I'm a little confused by this homomorphism:

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

If \\(\phi = \\{1 \mapsto 1, 2 \mapsto 1\\}\\), then \\(f(a_1 s_1 + a_2 s_2) = a_1 s_1 + a_2 a_1 = (a_1 + a_2) s_1\\), meaning this function doesn't _forget_ reagents, it just treats them as _interchangeable_. (But this does nicely disprove my hypothesis from [comment #1](https://forum.azimuthproject.org/discussion/comment/18335/#Comment_18335)!)

Thinking through my argument, it's only _accidentally_ correct for empty reaction networks. It's still true that we need minimal/maximal elements of \\(\mathbb{N}\\) to cover the forgotten reagents, since an infinity of \\(S\\)-complexes collapse onto the same \\(T\\)-complex. But the only element whose inverse image we need to worry about for computing \\(g(x)\\) is \\(x\\) itself.

(**EDIT:** Wait, I'm even doubting this now -- because no two of the collapsing infinity of \\(S\\)-complexes are related, we don't have a unique join.)

Also, I'm a little confused by this homomorphism:

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

If \\(\phi = \\{1 \mapsto 1, 2 \mapsto 1\\}\\), then \\(f(a_1 s_1 + a_2 s_2) = a_1 s_1 + a_2 a_1 = (a_1 + a_2) s_1\\), meaning this function doesn't _forget_ reagents, it just treats them as _interchangeable_. (But this does nicely disprove my hypothesis from [comment #1](https://forum.azimuthproject.org/discussion/comment/18335/#Comment_18335)!)