关于Lurie《高等拓扑学》§A.3.1中β_X,S定义的困惑：是否误用态射？

Great question—this is a super common point of confusion when working through Lurie's enriched adjunction framework, so let's break it down clearly.

First, short answer: your understanding is off the mark—the morphism $\operatorname{Map}(X, GY) \to \operatorname{Map}(FX, FGY)$ you're noticing isn't using a pre-existing $\mathbf{S}$-enriched structure on $F$. Instead, it's a natural consequence of the ordinary adjunction between $F$ and $G$, paired with the $\mathbf{S}$-enriched structure of the categories $\mathcal{C}$ and $\mathcal{D}$.

Let's unpack this step by step:

1. Recall the setup in §A.3.1
   We have ordinary adjoint functors $F: \mathcal{C} \to \mathcal{D}$ (left adjoint) and $G: \mathcal{D} \to \mathcal{C}$ (right adjoint), where both $\mathcal{C}$ and $\mathcal{D}$ are $\mathbf{S}$-enriched categories (here $\mathbf{S}$ denotes the category of simplicial sets). Crucially, $F$ is not assumed to be an $\mathbf{S}$-enriched functor at this stage.

2. How $\beta_{X,S}$ is constructed (without enriched $F$)
   The morphism $\beta_{X,S}: S \otimes FX \to F(S \otimes X)$ is defined via the Yoneda lemma, using natural isomorphisms from the adjunction and the definition of tensor products in $\mathbf{S}$-enriched categories:

   • By definition of the tensor $S \otimes (-)$, we have $\mathcal{D}(S \otimes FX, Z) \cong \operatorname{Fun}(S, \mathcal{D}(FX, Z))$ for any $Z \in \mathcal{D}$.
   • Using the ordinary adjunction $F \dashv G$, this becomes $\operatorname{Fun}(S, \mathcal{C}(X, GZ))$.
   • Again using the tensor definition, this is isomorphic to $\mathcal{C}(S \otimes X, GZ)$.
   • Finally, applying the adjunction once more, we get $\mathcal{D}(F(S \otimes X), Z)$.
     By Yoneda, these natural isomorphisms induce the morphism $\beta_{X,S}$.

3. Where does $\operatorname{Map}(X, GY) \to \operatorname{Map}(FX, FGY)$ come from?
   The morphism you're spotting isn't a structure on $F$, but a byproduct of the adjunction and the enriched structure of the categories:

   • For any $X \in \mathcal{C}$ and $Y \in \mathcal{D}$, the ordinary adjunction gives a natural simplicial set isomorphism $\mathcal{D}(FX, Y) \cong \mathcal{C}(X, GY)$. Since $\operatorname{Map}{\mathcal{D}}(FX, Y) = \mathcal{D}(FX, Y)$ and $\operatorname{Map}{\mathcal{C}}(X, GY) = \mathcal{C}(X, GY)$ (by definition of $\mathbf{S}$-enriched categories), this is an isomorphism of simplicial sets: $\operatorname{Map}{\mathcal{C}}(X, GY) \cong \operatorname{Map}{\mathcal{D}}(FX, Y)$.
   • Now, take the unit morphism of the adjunction $\eta: \text{id}{\mathcal{D}} \to FG$. Applying this to $Y$ gives $\eta_Y: Y \to FGY$, which induces a morphism $\operatorname{Map}{\mathcal{D}}(FX, Y) \to \operatorname{Map}_{\mathcal{D}}(FX, FGY)$ (via natural operations on simplicial mapping spaces).
   • Composing the isomorphism above with this induced morphism gives $\operatorname{Map}{\mathcal{C}}(X, GY) \to \operatorname{Map}{\mathcal{D}}(FX, FGY)$.

   The key point here is that this entire chain only uses:

   • The ordinary adjunction between $F$ and $G$,
   • The $\mathbf{S}$-enriched structure of $\mathcal{C}$ and $\mathcal{D}$ (i.e., their mapping spaces are simplicial sets),
   • Natural operations on simplicial mapping spaces (induced by morphisms in the category).

   It does not require $F$ to have a pre-defined $\mathbf{S}$-enriched functor structure (which would demand a compatible morphism $\operatorname{Map}{\mathcal{C}}(X, X') \to \operatorname{Map}{\mathcal{D}}(FX, FX')$ for all $X, X' \in \mathcal{C}$, satisfying enrichment axioms).

To wrap up: You were conflating a morphism induced by the adjunction and category enrichment with a built-in enriched functor structure on $F$. Lurie's definition avoids assuming $F$ is $\mathbf{S}$-enriched, and the mapping space morphism you noticed is a natural consequence of the given adjunction instead.

内容的提问来源于stack exchange，提问作者aaa