Fully faithful essentially surjective operad map

Proof. For every finite set \(I\), the fiber \(\Oo ^{\otimes }_I\) is equivalent to \(\Oo _{\lra {1}}^I\), and similarly for \(\Pp \). Under these equivalences, the functor induced by \(f^{\otimes }\) is the product of \(I\) copies of the underlying functor of \(f\), so it is essentially surjective. Thus \(f^{\otimes }\) is essentially surjective on objects.

To prove full faithfulness, let \(X=\{x_i\}_{i\in I}\) and \(Y=\{y_j\}_{j\in J}\) be objects of \(\Oo ^{\otimes }\). There is a commutative triangle

For a span \(\alpha \colon I\xleftarrow {u}K\xrightarrow {v}J\), the naturality in Lemma 14.1.6 identifies the induced map on fibers over \(\alpha \) with \[ \prod _{j\in J}\Oo (\{x_{u(k)}\}_{k\in v^{-1}(j)};y_j) \longrightarrow \prod _{j\in J}\Pp (\{f(x_{u(k)})\}_{k\in v^{-1}(j)};f(y_j)). \] This is an equivalence because \(f\) is fully faithful. Hence the top map in the triangle is an equivalence, since it is a map of animae over a fixed base which is an equivalence on every fiber. Thus \(f^{\otimes }\) is fully faithful and essentially surjective, and hence an equivalence over \(\Span (\Fin )\). □