 Solution

Let $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ forget the last partial unary operation. For an object $A$ of the nested partial unary operation category ${\mathcal{C}}_n$, setThe nesting rule implies that every earlier operation is defined and fixes such an $x$. To construct a free extension of nested partial unary operations, adjoin a disjoint infinite chain for each $x\in S_n(A)$:Keep the old operations on $A$. On each new chain put ${\alpha }_1(x,k)=(x,k+1)$; every old operation ${\alpha }_i$ with $2\le i\le n$ is undefined there, because the first operation has no fixed points there. Define the new operation only on $S_n(A)$, by ${\alpha }_{n+1}(x)=(x,0)$. These are exactly the fixed points of the old top operation, so the required domain rule is satisfied. The inclusion $A\hookrightarrow U_n{L}_{n}A$ preserves all old defined operations.

For a morphism $f:A\to U_{n}B$, preservation of the old operations makes $f(x)\in S_n(B)$ whenever $x\in S_n(A)$. Any extension to a morphism $\hat{f}:L_{n}A\to B$ must haveThe operation ${\beta }_{n+1}$ is defined at $f(x)$, and ${\beta }_1$ is total, so this formula exists. It preserves the new operation and the chain operation; there are no other defined operations on new points to check. It therefore gives the unique extension. Naturality and functoriality follow from uniqueness. ThusNotice the useful extra fact: the new top operation has no fixed points, since every value $(x,0)$ is new and every input $x$ is old. Hence freely adding any further operations adds no more points; those operations have empty domains.