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1. • CommentRowNumber1.
   • CommentAuthormetaweta
   • CommentTimeJul 14th 2016
   • PermaLink
   Author: metaweta
   Format: MarkdownItexAt "Definition as a 2-functor", the [article](https://ncatlab.org/nlab/show/2-category+equipped+with+proarrows) says > For a proarrow $H\colon B\to D$ and ordinary arrows $f\colon A\to B$ and $g\colon C\to D$, we write $H(g,f)$ for the composite $D(g,1) \circ H \circ B(1,f)$; it is a proarrow from $A$ to $C$. We also write $U_A$, $A(1,1)$, or simply $A$ for the identity proarrow $A\nrightarrow A$. In seems that the nLab has settled on the convention that a profunctor $H\colon B\nrightarrow D$ is a functor $H\colon D^{op} \times B \to Set$. Therefore, in the expressions $H(g, f)$ and $B(1, f)$ the morphism $f$ is in the covariant slot, which means that either $f$ should be a morphism out of $B$ or $H$ should be a profunctor into $A$. What am I missing?

   At “Definition as a 2-functor”, the article says

   For a proarrow $H\colon B\to D$ and ordinary arrows $f\colon A\to B$ and $g\colon C\to D$, we write $H(g,f)$ for the composite $D(g,1) \circ H \circ B(1,f)$; it is a proarrow from $A$ to $C$. We also write $U_A$, $A(1,1)$, or simply $A$ for the identity proarrow $A\nrightarrow A$.

   In seems that the nLab has settled on the convention that a profunctor $H\colon B\nrightarrow D$ is a functor $H\colon D^{op} \times B \to Set$. Therefore, in the expressions $H(g, f)$ and $B(1, f)$ the morphism $f$ is in the covariant slot, which means that either $f$ should be a morphism out of $B$ or $H$ should be a profunctor into $A$.

   What am I missing?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJul 14th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIt seems right to me. The covariant slot of $H$ is the category that it is "from" as a profunctor, and we are substituting a value of $f$ into that slot; so it's correct that $H$ is a profunctor from $B$ to something else, and that the values of $f$ belong to $B$, i.e. that $f$ is a morphism into $B$.

   It seems right to me. The covariant slot of $H$ is the category that it is “from” as a profunctor, and we are substituting a value of $f$ into that slot; so it’s correct that $H$ is a profunctor from $B$ to something else, and that the values of $f$ belong to $B$, i.e. that $f$ is a morphism into $B$.

3. • CommentRowNumber3.
   • CommentAuthormetaweta
   • CommentTimeJul 14th 2016
   • (edited Jul 14th 2016)
   • PermaLink
   Author: metaweta
   Format: MarkdownItexIn that case, $D$ and $g$ are the problem. Thinking of $H$ as a set of heteromorphisms from $B$ to $D$, we can precompose with $f$ to get a set of heteromorphisms from $A$ to $D$, but to postcompose, we need $g\colon D \to C$, not $g\colon C \to D$. That is, \[H(f\colon A\to B, g\colon D \to C)(h\colon B \to D) = g \circ h \circ f\colon A \to C.\]

   In that case, $D$ and $g$ are the problem. Thinking of $H$ as a set of heteromorphisms from $B$ to $D$, we can precompose with $f$ to get a set of heteromorphisms from $A$ to $D$, but to postcompose, we need $g\colon D \to C$, not $g\colon C \to D$. That is,

   $$H(f\colon A\to B, g\colon D \to C)(h\colon B \to D) = g \circ h \circ f\colon A \to C.$$
4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeJul 15th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex$f$ is not a morphism in $B$, it's a functor with codomain $B$, and similarly $g$ is not a morphism in $D$, it's a functor with codomain $D$. We're not describing the action of the functor $H$ on its elements/heteromorphisms; we're describing its *restriction* along a pair of functors.

   $f$ is not a morphism in $B$, it’s a functor with codomain $B$, and similarly $g$ is not a morphism in $D$, it’s a functor with codomain $D$. We’re not describing the action of the functor $H$ on its elements/heteromorphisms; we’re describing its restriction along a pair of functors.

5. • CommentRowNumber5.
   • CommentAuthorDavid_Corfield
   • CommentTimeJul 15th 2016
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexIt's like the multiplication of three matrices (the outer two being special in deriving from functions), if that helps. Or the composition of relations.

   It’s like the multiplication of three matrices (the outer two being special in deriving from functions), if that helps. Or the composition of relations.

6. • CommentRowNumber6.
   • CommentAuthormetaweta
   • CommentTimeJul 15th 2016
   • (edited Jul 15th 2016)
   • PermaLink
   Author: metaweta
   Format: MarkdownItexMike: Then shouldn't it be $D(1, g) \circ H \circ B(1,f)$, that is, $f$ followed by $H$ followed by the taking the preimage under $g$? The expression $D(g, 1)$ doesn't make sense; since $B$ is the *codomain* of $f$, having $g$ in the other slot means we'd want the *domain* of $g$. If we want the codomain of $g$, it needs to be in the same slot that $f$ is.

   Mike: Then shouldn’t it be $D(1, g) \circ H \circ B(1,f)$, that is, $f$ followed by $H$ followed by the taking the preimage under $g$? The expression $D(g, 1)$ doesn’t make sense; since $B$ is the codomain of $f$, having $g$ in the other slot means we’d want the domain of $g$. If we want the codomain of $g$, it needs to be in the same slot that $f$ is.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeJul 15th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThe $D$ in $D(g,1)$ represents the hom-functor of $D$, which is a profunctor from $D$ to $D$. Thus, $g$ being a functor with codomain $D$, it can be plugged into either slot of this profunctor.

   The $D$ in $D(g,1)$ represents the hom-functor of $D$, which is a profunctor from $D$ to $D$. Thus, $g$ being a functor with codomain $D$, it can be plugged into either slot of this profunctor.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJul 15th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexAlthough in the context of the quote in question, $D(1,g)$ and $D(g,1)$ are just *notations* for the two proarrows associated to the arrow $g$. The reason for using that notation is as I said: in the case of categories and profunctors, they are $(x,y) \mapsto Hom_D(x,g(y))$ and $(y,x) \mapsto Hom_D(g(y),x)$.

   Although in the context of the quote in question, $D(1,g)$ and $D(g,1)$ are just notations for the two proarrows associated to the arrow $g$. The reason for using that notation is as I said: in the case of categories and profunctors, they are $(x,y) \mapsto Hom_D(x,g(y))$ and $(y,x) \mapsto Hom_D(g(y),x)$.

9. • CommentRowNumber9.
   • CommentAuthormetaweta
   • CommentTimeJul 15th 2016
   • (edited Jul 15th 2016)
   • PermaLink
   Author: metaweta
   Format: MarkdownItexOK, so in that notation, both slots are covariant. I'll add a note to that effect. Thanks!

   OK, so in that notation, both slots are covariant. I’ll add a note to that effect. Thanks!

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeJul 15th 2016
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    Author: Mike Shulman
    Format: MarkdownItexI think I have some idea of what you mean, but I don't think "both slots are covariant" is really the correct way to say it. $H$ is still a functor from $D^{op}\times B$ to $Set$, so in that sense the first slot is contravariant. And if the question is what kind of a functor $H(g,f)$ is as a functor of $g$ and $f$, then it's a functor from $K(A,B)^{op}\times K(C,D)$ to $M(A,C)$, so in that sense also the first slot is contravariant.

    I think I have some idea of what you mean, but I don’t think “both slots are covariant” is really the correct way to say it. $H$ is still a functor from $D^{op}\times B$ to $Set$, so in that sense the first slot is contravariant. And if the question is what kind of a functor $H(g,f)$ is as a functor of $g$ and $f$, then it’s a functor from $K(A,B)^{op}\times K(C,D)$ to $M(A,C)$, so in that sense also the first slot is contravariant.