If "tensor" has an adjoint, is it automatically an "internal Hom"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:13:20Z http://mathoverflow.net/feeds/question/21382 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21382/if-tensor-has-an-adjoint-is-it-automatically-an-internal-hom If "tensor" has an adjoint, is it automatically an "internal Hom"? Theo Johnson-Freyd 2010-04-14T19:55:24Z 2010-05-25T21:46:52Z <p>Let <code>$\mathcal C,\otimes$</code> be a monoidal category, i.e. <code>$\otimes : \mathcal C \times \mathcal C \to \mathcal C$</code> is a functor, and there's a bit more structure and properties. Suppose that for each <code>$X \in \mathcal C$</code>, the functor <code>$X \otimes - : \mathcal C \to \mathcal C$</code> has a right adjoint. I will call this adjoint (unique up to canonical isomorphism of functors) <code>$\underline{\rm Hom}(X,-) : \mathcal C \to \mathcal C$</code>. By general abstract nonsense, <code>$\underline{\rm Hom}(X,-)$</code> is contravariant in <code>$X$</code>, and so defines a functor <code>$\underline{\rm Hom}: \mathcal C^{\rm op} \times \mathcal C \to \mathcal C$</code>. If <code>$1 \in \mathcal C$</code> is the monoidal unit, then <code>$\underline{\rm Hom}(1,-)$</code> is (naturally isomorphic to) the identity functor.</p> <p>Then there are canonically defined "evaluation" and "internal composition" maps, both of which I will denote by <code>$\bullet$</code>. Indeed, we define "evaluation" <code>$\bullet_{X,Y}: X\otimes \underline{\rm Hom}(X,Y) \to Y$</code> to be the map that corresponds to <code>${\rm id}: \underline{\rm Hom}(X,Y) \to \underline{\rm Hom}(X,Y)$</code> under the adjuntion. Then we define "composition" <code>$\bullet_{X,Y,Z}: \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$</code> to be the map that corresponds under the adjunction to <code>$\bullet_{Y,Z} \circ (\bullet_{X,Y} \otimes {\rm id}) : X \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to Z$</code>. (I have supressed all associators.)</p> <blockquote> <p><strong>Question:</strong> Is <code>$\bullet$</code> an associative multiplication? I.e. do we have necessarily equality of morphisms <code>$\bullet_{W,Y,Z} \circ (\bullet_{W,X,Y} \otimes {\rm id}) \overset ? = \bullet_{W,X,Z} \circ ({\rm id}\otimes \bullet_{X,Y,Z})$</code> of maps <code>$\underline{\rm Hom}(W,X) \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$</code>? If not, what extra conditions on <code>$\otimes$</code> are necessary/sufficient?</p> </blockquote> http://mathoverflow.net/questions/21382/if-tensor-has-an-adjoint-is-it-automatically-an-internal-hom/21385#21385 Answer by Evan Jenkins for If "tensor" has an adjoint, is it automatically an "internal Hom"? Evan Jenkins 2010-04-14T20:33:38Z 2010-04-14T20:33:38Z <p>It is associative. Consider the evaluation cube drawn <a href="http://math.uchicago.edu/~ejenkins/misc/internal-hom.pdf" rel="nofollow">here</a>. Four of the faces commute by definition of the composition map, and one by functoriality of the tensor product. The commutativity of these five faces implies that any of the maps $W \otimes \operatorname{Hom}(W, X) \otimes \operatorname{Hom}(X, Y) \otimes \operatorname{Hom}(Y, Z) \to Z$ are equal, so by adjunction, the two composites of compositions are equal.</p> http://mathoverflow.net/questions/21382/if-tensor-has-an-adjoint-is-it-automatically-an-internal-hom/25940#25940 Answer by Buschi Sergio for If "tensor" has an adjoint, is it automatically an "internal Hom"? Buschi Sergio 2010-05-25T21:46:52Z 2010-05-25T21:46:52Z <p>In S. Eilenberg and G. M. Kelly Closed categories, in Proc. C. O. C. A.. (La Jolla, 1965), </p> <p>These is a comprehensive study about Monoidal and Close structure on a category, and the relation and equivalence between these.</p>