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Todd Trimble
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This is the right Kan extension of $\hom_A: A \nrightarrow A$ along $K: A \nrightarrow B$ in the bicategory of profunctors. Which is to say that for every profunctor (aka bimodule) $L: B \nrightarrow A$ there is a natural bijection between morphisms $LK \to \hom_A$ and morphisms $L \to \mathcal{L}(K)$. Since $\hom_A$ is a unit $1_A$ in this bicategory, it may be more suggestive to write $\mathcal{L}(K) = Ran_K 1_A$.

Thus for $H: A \nrightarrow B$ and $K: B \nrightarrow C$, we have a canonical map $\mathcal{L}(K)K \to 1_B$, and whiskering on the right by $H$ and on the left by $\mathcal{L}(H)$, we get a composite

$$\mathcal{L}(H)\mathcal{L}(K)KH \to \mathcal{L}(H)1_B H \cong \mathcal{L}(H)H \to 1_A.$$

By the universal property of $\mathcal{L}(KH)$, we now obtain a canonical map $\mathcal{L}(H)\mathcal{L}(K) \to \mathcal{L}(KH)$.

But this map is typically not an isomorphism. The simplest type of example that comes to mind is just the classical case of bimodules over rings, where for a right-$B$ left-$C$ bimodule $K$, we have $\mathcal{L}(K) = \hom_A(K, B)$ regarded as a right-$C$ left-$B$ bimodule. We have in this situation a canonical map

$$\hom_A(H, A) \otimes_B \hom_B(K, B) \to \hom_A(K \otimes_B H, A)$$

but normally this won't be an isomorphism. Indeed, even in the humble case of vector spaces over a ground field $k$, the canonical map $V^\ast \otimes_k W^\ast \to (V \otimes W)^\ast$ isn't generally an isomorphism. Of course we do get an isomorphism here in some special cases, such as if $K$ is finitely generated projective over $B$. More abstractly, this is the situation where $K$ has a left adjoint bimodule, and here we may recall that in a 2-category or bicategory, if an arrow $K: B \nrightarrow C$ has a left adjoint $L$, then it is necessarily $L = Ran_K 1_B$, i.e., $L = \mathcal{L}(K)$ in our situation. In that case, the asserted inverse $\mathcal{L}(KH) \to \mathcal{L}(H)\mathcal{L}(K)$ of the canonical map is mated (by the adjunction $\mathcal{L}(K) \dashv K$) to an arrow $\mathcal{L}(KH)K \to \mathcal{L}(H)$, which in turn is mated to the canonical arrow $\mathcal{L}(KH)KH \to 1_A$ using the definition of right Kan extension.

[For the bicategory of profunctors or bimodules, such right adjoints $K$ are induced by functors $F: C \to \bar{B}$ where $\bar{B}$ is the Cauchy completion of $B$ (which we may think of as analogous to the category of finitely generated projective $B$-modules); specifically, $K(b, c) = \bar{B}(b, Fc)$ and its left adjoint is given by $\mathcal{L}(K)(c, b) = \bar{B}(Fc, b)$.]

By similar reasoning as above, one may calculate that if $K: B \nrightarrow C$ has a left adjoint $L$ (in a bicategory $\mathbf{B}$), then for every $F: B \nrightarrow A$ we have an isomorphism $Ran_F(K) \cong K \circ Ran_F(1_B)$, assuming these right Kan extensions exist. This is purely formal of course: for every $H: A \nrightarrow C$ we have isomorphisms

$$\mathbf{B}(A, C)(H, Ran_F(K)) \cong \mathbf{B}(B, C)(HF, K) \cong \mathbf{B}(B, B)(LHF, 1_B) \cong \mathbf{B}(A, B)(LH, Ran_F(1_B)) \cong \mathbf{B}(A, C)(H, K Ran_F(1_B)).$$

I'm not sure what could be said about interaction with the "tautological dualizer" (where now are viewing $\textbf{Prof}$ as a compact closed bicategory): essentially all of the above has to do with the bicategory structure, not the monoidal bicategory structure.

This is the right Kan extension of $\hom_A: A \nrightarrow A$ along $K: A \nrightarrow B$ in the bicategory of profunctors. Which is to say that for every profunctor (aka bimodule) $L: B \nrightarrow A$ there is a natural bijection between morphisms $LK \to \hom_A$ and morphisms $L \to \mathcal{L}(K)$. Since $\hom_A$ is a unit $1_A$ in this bicategory, it may be more suggestive to write $\mathcal{L}(K) = Ran_K 1_A$.

Thus for $H: A \nrightarrow B$ and $K: B \nrightarrow C$, we have a canonical map $\mathcal{L}(K)K \to 1_B$, and whiskering on the right by $H$ and on the left by $\mathcal{L}(H)$, we get a composite

$$\mathcal{L}(H)\mathcal{L}(K)KH \to \mathcal{L}(H)1_B H \cong \mathcal{L}(H)H \to 1_A.$$

By the universal property of $\mathcal{L}(KH)$, we now obtain a canonical map $\mathcal{L}(H)\mathcal{L}(K) \to \mathcal{L}(KH)$.

But this map is typically not an isomorphism. The simplest type of example that comes to mind is just the classical case of bimodules over rings, where for a right-$B$ left-$C$ bimodule $K$, we have $\mathcal{L}(K) = \hom_A(K, B)$ regarded as a right-$C$ left-$B$ bimodule. We have in this situation a canonical map

$$\hom_A(H, A) \otimes_B \hom_B(K, B) \to \hom_A(K \otimes_B H, A)$$

but normally this won't be an isomorphism. Indeed, even in the humble case of vector spaces over a ground field $k$, the canonical map $V^\ast \otimes_k W^\ast \to (V \otimes W)^\ast$ isn't generally an isomorphism. Of course we do get an isomorphism here in some special cases, such as if $K$ is finitely generated projective over $B$. More abstractly, this is the situation where $K$ has a left adjoint bimodule, and here we may recall that in a 2-category or bicategory, if an arrow $K: B \nrightarrow C$ has a left adjoint $L$, then it is necessarily $L = Ran_K 1_B$, i.e., $L = \mathcal{L}(K)$ in our situation. In that case, the asserted inverse $\mathcal{L}(KH) \to \mathcal{L}(H)\mathcal{L}(K)$ of the canonical map is mated (by the adjunction $\mathcal{L}(K) \dashv K$) to an arrow $\mathcal{L}(KH)K \to \mathcal{L}(H)$, which in turn is mated to the canonical arrow $\mathcal{L}(KH)KH \to 1_A$ using the definition of right Kan extension.

[For the bicategory of profunctors or bimodules, such right adjoints $K$ are induced by functors $F: C \to \bar{B}$ where $\bar{B}$ is the Cauchy completion of $B$ (which we may think of as analogous to the category of finitely generated projective $B$-modules); specifically, $K(b, c) = \bar{B}(b, Fc)$ and its left adjoint is given by $\mathcal{L}(K)(c, b) = \bar{B}(Fc, b)$.]

By similar reasoning as above, one may calculate that if $K: B \nrightarrow C$ has a left adjoint $L$ (in a bicategory $\mathbf{B}$), then for every $F: B \nrightarrow A$ we have an isomorphism $Ran_F(K) \cong K \circ Ran_F(1_B)$, assuming these right Kan extensions exist. This is purely formal of course: for every $H: A \nrightarrow C$ we have isomorphisms

$$\mathbf{B}(A, C)(H, Ran_F(K)) \cong \mathbf{B}(B, C)(HF, K) \cong \mathbf{B}(B, B)(LHF, 1_B) \cong \mathbf{B}(A, B)(LH, Ran_F(1_B)) \cong \mathbf{B}(A, C)(H, K Ran_F(1_B)).$$

I'm not sure what could be said about interaction with the "tautological dualizer" (where now are viewing $\textbf{Prof}$ as a compact closed bicategory: essentially all of the above has to do with the bicategory structure, not the monoidal bicategory structure.

This is the right Kan extension of $\hom_A: A \nrightarrow A$ along $K: A \nrightarrow B$ in the bicategory of profunctors. Which is to say that for every profunctor (aka bimodule) $L: B \nrightarrow A$ there is a natural bijection between morphisms $LK \to \hom_A$ and morphisms $L \to \mathcal{L}(K)$. Since $\hom_A$ is a unit $1_A$ in this bicategory, it may be more suggestive to write $\mathcal{L}(K) = Ran_K 1_A$.

Thus for $H: A \nrightarrow B$ and $K: B \nrightarrow C$, we have a canonical map $\mathcal{L}(K)K \to 1_B$, and whiskering on the right by $H$ and on the left by $\mathcal{L}(H)$, we get a composite

$$\mathcal{L}(H)\mathcal{L}(K)KH \to \mathcal{L}(H)1_B H \cong \mathcal{L}(H)H \to 1_A.$$

By the universal property of $\mathcal{L}(KH)$, we now obtain a canonical map $\mathcal{L}(H)\mathcal{L}(K) \to \mathcal{L}(KH)$.

But this map is typically not an isomorphism. The simplest type of example that comes to mind is just the classical case of bimodules over rings, where for a right-$B$ left-$C$ bimodule $K$, we have $\mathcal{L}(K) = \hom_A(K, B)$ regarded as a right-$C$ left-$B$ bimodule. We have in this situation a canonical map

$$\hom_A(H, A) \otimes_B \hom_B(K, B) \to \hom_A(K \otimes_B H, A)$$

but normally this won't be an isomorphism. Indeed, even in the humble case of vector spaces over a ground field $k$, the canonical map $V^\ast \otimes_k W^\ast \to (V \otimes W)^\ast$ isn't generally an isomorphism. Of course we do get an isomorphism here in some special cases, such as if $K$ is finitely generated projective over $B$. More abstractly, this is the situation where $K$ has a left adjoint bimodule, and here we may recall that in a 2-category or bicategory, if an arrow $K: B \nrightarrow C$ has a left adjoint $L$, then it is necessarily $L = Ran_K 1_B$, i.e., $L = \mathcal{L}(K)$ in our situation. In that case, the asserted inverse $\mathcal{L}(KH) \to \mathcal{L}(H)\mathcal{L}(K)$ of the canonical map is mated (by the adjunction $\mathcal{L}(K) \dashv K$) to an arrow $\mathcal{L}(KH)K \to \mathcal{L}(H)$, which in turn is mated to the canonical arrow $\mathcal{L}(KH)KH \to 1_A$ using the definition of right Kan extension.

[For the bicategory of profunctors or bimodules, such right adjoints $K$ are induced by functors $F: C \to \bar{B}$ where $\bar{B}$ is the Cauchy completion of $B$ (which we may think of as analogous to the category of finitely generated projective $B$-modules); specifically, $K(b, c) = \bar{B}(b, Fc)$ and its left adjoint is given by $\mathcal{L}(K)(c, b) = \bar{B}(Fc, b)$.]

By similar reasoning as above, one may calculate that if $K: B \nrightarrow C$ has a left adjoint $L$ (in a bicategory $\mathbf{B}$), then for every $F: B \nrightarrow A$ we have an isomorphism $Ran_F(K) \cong K \circ Ran_F(1_B)$, assuming these right Kan extensions exist. This is purely formal of course: for every $H: A \nrightarrow C$ we have isomorphisms

$$\mathbf{B}(A, C)(H, Ran_F(K)) \cong \mathbf{B}(B, C)(HF, K) \cong \mathbf{B}(B, B)(LHF, 1_B) \cong \mathbf{B}(A, B)(LH, Ran_F(1_B)) \cong \mathbf{B}(A, C)(H, K Ran_F(1_B)).$$

I'm not sure what could be said about interaction with the "tautological dualizer" (where now are viewing $\textbf{Prof}$ as a compact closed bicategory): essentially all of the above has to do with the bicategory structure, not the monoidal bicategory structure.

Source Link
Todd Trimble
  • 53.3k
  • 6
  • 205
  • 322

This is the right Kan extension of $\hom_A: A \nrightarrow A$ along $K: A \nrightarrow B$ in the bicategory of profunctors. Which is to say that for every profunctor (aka bimodule) $L: B \nrightarrow A$ there is a natural bijection between morphisms $LK \to \hom_A$ and morphisms $L \to \mathcal{L}(K)$. Since $\hom_A$ is a unit $1_A$ in this bicategory, it may be more suggestive to write $\mathcal{L}(K) = Ran_K 1_A$.

Thus for $H: A \nrightarrow B$ and $K: B \nrightarrow C$, we have a canonical map $\mathcal{L}(K)K \to 1_B$, and whiskering on the right by $H$ and on the left by $\mathcal{L}(H)$, we get a composite

$$\mathcal{L}(H)\mathcal{L}(K)KH \to \mathcal{L}(H)1_B H \cong \mathcal{L}(H)H \to 1_A.$$

By the universal property of $\mathcal{L}(KH)$, we now obtain a canonical map $\mathcal{L}(H)\mathcal{L}(K) \to \mathcal{L}(KH)$.

But this map is typically not an isomorphism. The simplest type of example that comes to mind is just the classical case of bimodules over rings, where for a right-$B$ left-$C$ bimodule $K$, we have $\mathcal{L}(K) = \hom_A(K, B)$ regarded as a right-$C$ left-$B$ bimodule. We have in this situation a canonical map

$$\hom_A(H, A) \otimes_B \hom_B(K, B) \to \hom_A(K \otimes_B H, A)$$

but normally this won't be an isomorphism. Indeed, even in the humble case of vector spaces over a ground field $k$, the canonical map $V^\ast \otimes_k W^\ast \to (V \otimes W)^\ast$ isn't generally an isomorphism. Of course we do get an isomorphism here in some special cases, such as if $K$ is finitely generated projective over $B$. More abstractly, this is the situation where $K$ has a left adjoint bimodule, and here we may recall that in a 2-category or bicategory, if an arrow $K: B \nrightarrow C$ has a left adjoint $L$, then it is necessarily $L = Ran_K 1_B$, i.e., $L = \mathcal{L}(K)$ in our situation. In that case, the asserted inverse $\mathcal{L}(KH) \to \mathcal{L}(H)\mathcal{L}(K)$ of the canonical map is mated (by the adjunction $\mathcal{L}(K) \dashv K$) to an arrow $\mathcal{L}(KH)K \to \mathcal{L}(H)$, which in turn is mated to the canonical arrow $\mathcal{L}(KH)KH \to 1_A$ using the definition of right Kan extension.

[For the bicategory of profunctors or bimodules, such right adjoints $K$ are induced by functors $F: C \to \bar{B}$ where $\bar{B}$ is the Cauchy completion of $B$ (which we may think of as analogous to the category of finitely generated projective $B$-modules); specifically, $K(b, c) = \bar{B}(b, Fc)$ and its left adjoint is given by $\mathcal{L}(K)(c, b) = \bar{B}(Fc, b)$.]

By similar reasoning as above, one may calculate that if $K: B \nrightarrow C$ has a left adjoint $L$ (in a bicategory $\mathbf{B}$), then for every $F: B \nrightarrow A$ we have an isomorphism $Ran_F(K) \cong K \circ Ran_F(1_B)$, assuming these right Kan extensions exist. This is purely formal of course: for every $H: A \nrightarrow C$ we have isomorphisms

$$\mathbf{B}(A, C)(H, Ran_F(K)) \cong \mathbf{B}(B, C)(HF, K) \cong \mathbf{B}(B, B)(LHF, 1_B) \cong \mathbf{B}(A, B)(LH, Ran_F(1_B)) \cong \mathbf{B}(A, C)(H, K Ran_F(1_B)).$$

I'm not sure what could be said about interaction with the "tautological dualizer" (where now are viewing $\textbf{Prof}$ as a compact closed bicategory: essentially all of the above has to do with the bicategory structure, not the monoidal bicategory structure.