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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 5th 2010
   • (edited Aug 5th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexI asked this question over at MO, and I figured if anyone here wants to answer it, here ya go: [Click](http://mathoverflow.net/questions/34525/a-better-way-to-compute-the-mapping-spaces-of-the-category-of-spans-in-an-enriche) 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 $X^{Sp}$ 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: $Map_C(a,b)$ is the object of $V$ representing the functor $M_{ab}(\gamma):= Hom_C(\gamma \otimes a, b)$ (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: $$Map_C(a,b)=Map_X(A,B)\underset{Map_X(A,B'\times B'')}{\times} (Map_X(A',B')\times Map_X(A'',B''))$$ Where $a=A'\leftarrow A \to A''$ and $b=B'\leftarrow B \to B''$. To prove that these two descriptions are equivalent, I applied Yoneda's lemma to the second definition of $Map_C(a,b)$, which gives us $$Hom_V(Q,Map_C(a,b))=Hom_X(Q\otimes A, B)\underset{Hom_X(Q\otimes A,B'\times B'')}{\times}(Hom_X(Q\otimes A',B')\times Hom_X(Q\otimes A'',B''))$$ Which by the ordinary fiber product in the category of sets is precisely the set of triplets of arrows $(Q\otimes A\to B,(Q\otimes A'\to B',Q\otimes A''\to B''))$ 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?

   I asked this question over at MO, and I figured if anyone here wants to answer it, here ya go:

   Click

   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 $X^{Sp}$ 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: $Map_C(a,b)$ is the object of $V$ representing the functor $M_{ab}(\gamma):= Hom_C(\gamma \otimes a, b)$ (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:

   $$Map_C(a,b)=Map_X(A,B)\underset{Map_X(A,B'\times B'')}{\times} (Map_X(A',B')\times Map_X(A'',B''))$$

   Where $a=A'\leftarrow A \to A''$ and $b=B'\leftarrow B \to B''$.

   To prove that these two descriptions are equivalent, I applied Yoneda’s lemma to the second definition of $Map_C(a,b)$, which gives us

   $$Hom_V(Q,Map_C(a,b))=Hom_X(Q\otimes A, B)\underset{Hom_X(Q\otimes A,B'\times B'')}{\times}(Hom_X(Q\otimes A',B')\times Hom_X(Q\otimes A'',B''))$$

   Which by the ordinary fiber product in the category of sets is precisely the set of triplets of arrows $(Q\otimes A\to B,(Q\otimes A'\to B',Q\otimes A''\to B''))$ 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?