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This is probably an easy question. Let C be a category with (finite) products. An internal hom in C category is an object uhom(X, Z) which represents the functor:

Y |-----> hom(Y x X, Z)

here "uhom" is for "underlined hom" as that is how it is commonly denoted. Many example of categories with internal homs satisfy an a priori stronger for of adjunction:

uhom(Y x X. Z) = uhom(Y, uhom(X,Z))

Is this automatic for categories with internal homs? Is there an easily understood counter example?

(*) This might not be the most general/best definition of internal hom, but it is valid for many examples.

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up vote 8 down vote accepted

I think both uhom(Y × X, Z) and uhom(Y, uhom(X, Z)) represent the same functor: hom(W, uhom(Y × X, Z)) = hom(W × Y × X, Z) hom(W, uhom(Y, uhom(X, Z))) = hom(W × Y, uhom(X, Z)) = hom(W × Y × X, Z), hence by Yoneda lemma they are isomorphic.

As a side remark, the right notion of inner hom in the absence of monoidal structure is the notion of closed category:

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right. That was embarrassingly easy. Thanks. – Chris Schommer-Pries Oct 29 '09 at 13:09

Isn't this simply Yoneda given the associativity of products?

Hom(T,uHom(Y x X,Z)) = Hom(T x Y x X,Z) Hom(T x Y, uHom(X,Z)) = Hom(T,uHom(Y,uHom(X,Z)) for any test object T.

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As Dimitri ans bhargav said, this is automatic for categories with internal homs. And you don't need x to be the product: you have the same result for any closed monoidal category (in your case it is cartesian closed). You can find it in "Basic concepts of enriched category theory", by G.M. Kelly, page 14, which you can get for free here: .

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