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.

**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.