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I asked this question over at MO, and I figured if anyone here wants to answer it, here ya go:
Full question:
Let X be a tensored and cotensored V-category, where V is a fixed complete, cocomplete, closed symmetric monoidal category.
Define C:=Span(X) to be the category of spans in X (this is the functor category XSp where Sp is the walking span). We notice that C is automatically “tensored” over V (by computing the tensor product pointwise). Then C has a natural V-enriched structure given as follows: MapC(a,b) is the object of V representing the functor Mab(γ):=HomC(γ⊗a,b) (such an object exists by the adjoint functor theorem and since the tensor product is cocontinuous).
We can give another description of the mapping space as:
MapC(a,b)=MapX(A,B)×MapX(A,B′×B″)(MapX(A′,B′)×MapX(A″,B″))Where a=A′←A→A″ and b=B′←B→B″.
To prove that these two descriptions are equivalent, I applied Yoneda’s lemma to the second definition of MapC(a,b), which gives us
HomV(Q,MapC(a,b))=HomX(Q⊗A,B)×HomX(Q⊗A,B′×B″)(HomX(Q⊗A′,B′)×HomX(Q⊗A″,B″))Which by the ordinary fiber product in the category of sets is precisely the set of triplets of arrows (Q⊗A→B,(Q⊗A′→B′,Q⊗A″→B″)) giving the commutativity of the natural transformation diagram in X. This construction is obviously functorial in Q for fixed a and b.
Surely there must be a better way to do this, presumably without relying so heavily on the definition of the fiber product in the category of sets. What does such a proof look like? I assume there must be a simpler proof, because this fact was asserted as though it were trivial in a book I’m reading.
Question: What’s a slicker way to prove that the two definitions are equivalent?
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