First, let's establish that the general form of a complex is a function \$$x : S \to \mathbb{N}\$$ giving quantities for each resource, where \$$f\$$ has finite support (only finitely many values are nonzero). When \$$S\$$ is finite, this is just a tuple. In this setting, the monoidal monotone \$$f\$$ that forgets some elements of \$$S\$$ sends \$$x\$$ to its restriction on \$$T\$$.

**Puzzle 84:** Let's consider what a right adjoint \$$f \dashv g\$$ should look like. For complexes \$$x \in \mathbb{N}[S]\$$ and \$$y \in \mathbb{N}[T]\$$, we must have the property \$$f(x) \le y \Leftrightarrow x \le g(y)\$$. For any particular \$$T\$$-complex \$$y\$$, the set of \$$T\$$-complexes less than it are all those with no more \$$T\$$-resources than \$$y\$$. The set of \$$S\$$-complexes that map into this set have the same restriction on \$$T\$$-resources, but _no_ restriction on \$$(S \setminus T)\$$-resources. Requiring \$$g(y)\$$ to be greater than or equal to every one of these \$$S\$$-complexes is impossible, because there is no greatest natural number.

Thus, \$$f\$$ has no right adjoint.

**Puzzle 85:** Here, we need the property \$$g(x) \le y \Leftrightarrow x \le f(y)\$$, where \$$x \in \mathbb{N}[T]\$$ and \$$y \in \mathbb{N}[S]\$$. For any particular \$$T\$$-complex \$$x\$$, the set of \$$T\$$-complexes greater than it are all those with no more \$$T\$$-resources than \$$x\$$. The set of \$$S\$$-complexes that map into this set have the same restriction on \$$T\$$-resources, but _no_ restriction on \$$S \setminus T\$$-resources. Thus, in order to be less than or equal to every such \$$S\$$-complex, \$$g(x)\$$ must assign \$$0\$$ to every \$$S \setminus T\$$-resource:

\$g(x)(r) = \begin{cases} x(r) & \text{when } r \in T \\\\ 0 & \text{otherwise} \end{cases}\$

Finally, we know that \$$g\$$ is strict monoidal monotone. To put it simply, 0 + 0 is still 0, and the \$$T\$$-resource values are unchanged by \$$g\$$.

As an addendum, if \$$S \setminus T\$$ is infinite, then there is technically a restriction on \$$S\$$-complexes -- namely that it must have finite support over \$$S \setminus T\$$ resources. But the proof doesn't change if we explicitly discard those -- there's always a finitely supported complex (or a set of them) that we'd have to consider anyway which does its job.