When Bimodules Become Morphisms: The Morita Bicategory of R-Algebras

1. Algebra homomorphisms are special cases of bimodule 1-morphisms2. Morita equivalence becomes equivalence of objects3. Module Categories as Hom-Categories from the Unit ObjectThe Representable PseudofunctorMorita Equivalence Implies Equivalence of Module Categories

Let us denote the Morita bicategory by ${\mathbf{Alg}}_R$$\mathbf{Alg}_R$. We adopt the direction convention used above:

• The objects are unital associative $R$$R$-algebras $A,B,\dots$$A,B,\ldots$.

• A 1-morphism from $A$$A$ to $B$$B$ is an $A$$A$-$B$$B$-bimodule ${}_A{M}_B$${}_A M_B$.

• A 2-morphism $M\Rightarrow M^{\prime }$$M\Rightarrow M'$ is a homomorphism of $A$$A$-$B$$B$-bimodules.

Suppose that we have 1-morphisms

$${}_A{M}_B:A⟶B$$

and

$${}_B{N}_C:B⟶C.$$

Their composite is defined by

$$N\circ M:=M{\otimes }_{B}N,$$

which is naturally an $A$$A$-$C$$C$-bimodule. The balancing relation

$$mb\otimes n=m\otimes bn$$

expresses the fact that, when composing through the intermediate algebra $B$$B$, the right $B$$B$-action on $M$$M$ must be identified with the left $B$$B$-action on $N$$N$.

The identity 1-morphism on $A$$A$ is the regular bimodule

$$1_A={}_A{A}_A,$$

because there are natural isomorphisms

$$A{\otimes }_{A}M\cong M$$

and

$$M{\otimes }_{B}B\cong M.$$

Strictly speaking, this construction usually gives a bicategory rather than a strict 2-category, since

$$(M{\otimes }_{B}N){\otimes }_{C}P\cong M{\otimes }_B(N{\otimes }_{C}P)$$

only by a canonical natural isomorphism, rather than by literal equality.

What makes this construction especially remarkable is the following.

1. Algebra homomorphisms are special cases of bimodule 1-morphisms

Given an algebra homomorphism

$$f:A⟶B,$$

we may regard $B$$B$ as an $A$$A$-$B$$B$-bimodule

$${}_f{B}_B,$$

where the left $A$$A$-action is defined by

$$a\cdot b:=f(a)b.$$

Thus every ordinary algebra homomorphism determines a 1-morphism in the Morita bicategory. However, bimodules are considerably more general than algebra homomorphisms. The Morita bicategory enlarges the notion of a map between algebras into a notion of generalized morphism.

This is analogous to the inclusions

$$\text{functions}\subseteq \text{relations}$$

and

$$\text{maps}\subseteq \text{correspondences}.$$

In this sense, bimodules play the role of correspondences in the algebraic world.

2. Morita equivalence becomes equivalence of objects

Two algebras $A$$A$ and $B$$B$ are Morita equivalent precisely when there exist bimodules

$${}_A{P}_B$$

and

$${}_B{Q}_A$$

together with bimodule isomorphisms

$$P{\otimes }_{B}Q\cong A$$

and

$$Q{\otimes }_{A}P\cong B.$$

Equivalently, $P:A\to B$$P:A\to B$ and $Q:B\to A$$Q:B\to A$ are mutually inverse 1-morphisms up to invertible 2-morphisms.

Therefore, Morita equivalence is not an additional equivalence relation artificially imposed on algebras. It is exactly the intrinsic notion of equivalence between objects in the Morita bicategory.

For example,

$$R{\simeq }_{\mathrm{Mor}}{M}_n(R).$$

The algebras $R$$R$ and $M_n(R)$$M_n(R)$ are generally not isomorphic in the ordinary category of $R$$R$-algebras, but they are equivalent objects in the Morita bicategory. They should be regarded as equivalent because they have equivalent module categories:

$$\mathrm{Mod}\text{-}R\simeq \mathrm{Mod}\text{-}{M}_n(R).$$

3. Module Categories as Hom-Categories from the Unit Object

The Morita bicategory ${\mathbf{Alg}}_R$$\mathbf{Alg}_R$ carries a symmetric monoidal structure whose tensor product on objects is

$$A⊠B:=A{\otimes }_{R}B.$$

Its monoidal unit is the ground ring $R$$R$, since there are canonical isomorphisms

$$R{\otimes }_{R}A\cong A$$

and

$$A{\otimes }_{R}R\cong A.$$

This monoidal unit should not be confused with the identity 1-morphism on an algebra $A$$A$. The latter is the regular bimodule

$$1_A={}_A{A}_A.$$

We now examine the hom-category from the monoidal unit $R$$R$ to an algebra $A$$A$:

$${\mathbf{Alg}}_R(R,A).$$

Its objects are $R$$R$-$A$$A$-bimodules, and its morphisms are homomorphisms of $R$$R$-$A$$A$-bimodules. In general, an $A$$A$-$B$$B$-bimodule is equivalently a left module over the algebra

$$A{\otimes }_R{B}^{op}.$$

Indeed, the corresponding action is defined by

$$(a\otimes b^{op})\cdot m:=amb.$$

Specializing to an $R$$R$-$A$$A$-bimodule, we obtain

$${}_R{\mathbf{Bimod}}_A\simeq (R{\otimes }_R{A}^{op})\text{-}\mathbf{Mod}.$$

Since

$$R{\otimes }_R{A}^{op}\cong A^{op},$$

we have

$${}_R{\mathbf{Bimod}}_A\simeq A^{op}\text{-}\mathbf{Mod}.$$

Finally, a left $A^{op}$$A^{op}$-module is the same thing as a right $A$$A$-module. Therefore,

$$\boxed{{\displaystyle {\mathbf{Alg}}_R(R,A)\simeq \mathrm{Mod}\text{-}A.}}$$

Thus the right $A$$A$-modules can be interpreted as the 1-morphisms from the monoidal unit $R$$R$ to the algebra $A$$A$.

In this sense, the modules over $A$$A$ are the categorified points of $A$$A$ in the Morita bicategory. Instead of obtaining a set of points, we obtain an entire category consisting of modules and module homomorphisms.

The Representable Pseudofunctor

Fixing the source object $R$$R$ gives a representable pseudofunctor

$${\mathbf{Alg}}_R(R,-):{\mathbf{Alg}}_R⟶\mathbf{Cat}.$$

On objects, it sends an algebra $A$$A$ to its category of right modules:

$$A⟼{\mathbf{Alg}}_R(R,A)\simeq \mathrm{Mod}\text{-}A.$$

Now let

$${}_A{M}_B:A⟶B$$

be a 1-morphism in the Morita bicategory. Postcomposition with $M$$M$ gives a functor

$${\mathbf{Alg}}_R(R,A)⟶{\mathbf{Alg}}_R(R,B).$$

Under the identification with module categories, this is precisely the tensor functor

$${\mathrm{\Phi }}_M:=-{\otimes }_{A}M:\mathrm{Mod}\text{-}A⟶\mathrm{Mod}\text{-}B.$$

Explicitly, a right $A$$A$-module $X$$X$, regarded as a 1-morphism

$$X:R⟶A,$$

is sent to the composite

$$R\stackrel{X}{\to }A\stackrel{M}{\to }B,$$

which is represented by the right $B$$B$-module

$$X{\otimes }_{A}M.$$

Similarly, a 2-morphism of bimodules

$$\phi :M⟹M^{\prime }$$

induces a natural transformation

$${\mathrm{\Phi }}_{\phi }:-{\otimes }_{A}M⟹-{\otimes }_A{M}^{\prime },$$

whose component at a right $A$$A$-module $X$$X$ is

$$({\mathrm{\Phi }}_{\phi }{)}_X={\mathrm{id}}_X{\otimes }_A\phi .$$

This construction preserves composition only up to canonical natural isomorphism. If

$${}_A{M}_B:A⟶B$$

and

$${}_B{N}_C:B⟶C,$$

then

$$\begin{array}{rl}{\mathrm{\Phi }}_N{\mathrm{\Phi }}_M(X)& =(X{\otimes }_{A}M){\otimes }_{B}N\\ & \cong X{\otimes }_A(M{\otimes }_{B}N)\\ & ={\mathrm{\Phi }}_{N\circ M}(X).\end{array}$$

Here the composite 1-morphism is

$$N\circ M=M{\otimes }_{B}N.$$

The appearance of the canonical associativity isomorphism explains why ${\mathbf{Alg}}_R(R,-)$$\mathbf{Alg}_R(R,-)$ is naturally a pseudofunctor rather than a strict 2-functor.

Morita Equivalence Implies Equivalence of Module Categories

Suppose that $A$$A$ and $B$$B$ are equivalent objects in the Morita bicategory. This means that there exist bimodules

$${}_A{M}_B:A⟶B$$

and

$${}_B{N}_A:B⟶A$$

together with bimodule isomorphisms

$$M{\otimes }_{B}N\cong A$$

and

$$N{\otimes }_{A}M\cong B.$$

The corresponding tensor functors are

$$F:=-{\otimes }_{A}M:\mathrm{Mod}\text{-}A⟶\mathrm{Mod}\text{-}B$$

and

$$G:=-{\otimes }_{B}N:\mathrm{Mod}\text{-}B⟶\mathrm{Mod}\text{-}A.$$

For every right $A$$A$-module $X$$X$, we have natural isomorphisms

$$\begin{array}{rl}GF(X)& =(X{\otimes }_{A}M){\otimes }_{B}N\\ & \cong X{\otimes }_A(M{\otimes }_{B}N)\\ & \cong X{\otimes }_{A}A\\ & \cong X.\end{array}$$

Therefore,

$$GF\cong {\mathrm{id}}_{\mathrm{Mod}\text{-}A}.$$

Similarly, for every right $B$$B$-module $Y$$Y$,

$$\begin{array}{rl}FG(Y)& =(Y{\otimes }_{B}N){\otimes }_{A}M\\ & \cong Y{\otimes }_B(N{\otimes }_{A}M)\\ & \cong Y{\otimes }_{B}B\\ & \cong Y.\end{array}$$

Hence,

$$FG\cong {\mathrm{id}}_{\mathrm{Mod}\text{-}B}.$$

Consequently,

$$\boxed{{\displaystyle A{\simeq }_{\mathrm{Mor}}B⟹\mathrm{Mod}\text{-}A\simeq \mathrm{Mod}\text{-}B.}}$$

This implication is a formal consequence of pseudofunctoriality: every pseudofunctor sends equivalent objects to equivalent categories. More generally, for any bicategory $\mathcal{B}$$\mathcal B$ and any fixed object $X$$X$, an equivalence

$$A\simeq B$$

induces an equivalence

$$\mathcal{B}(X,A)\simeq \mathcal{B}(X,B).$$

In the present situation, we take $X=R$$X=R$.

A Seven-Term Adjoint Chain in the Arrow Category

A direct proof with every relevant commutative square drawn explicitly.

Let $\mathcal A$ be an abelian category. Its arrow category $\mathcal A^{[1]}$ is the functor category associated with

$$ [1]=(0\longrightarrow 1). $$

It is also often denoted by $\mathcal A^2$. An object of $\mathcal A^{[1]}$ is simply a morphism

$$ f:X_0\longrightarrow X_1 $$

in $\mathcal A$.

A morphism $$ (a,b):f\longrightarrow g $$ from $f:X_0\to X_1$ to $g:Y_0\to Y_1$ is a commutative square, equivalently a pair of morphisms

$$ a:X_0\longrightarrow Y_0, \qquad b:X_1\longrightarrow Y_1 $$

satisfying $ga=bf$. In diagrammatic form:

A morphism in the arrow category.

$$ \begin{CD} X_0 @>{a}>> Y_0 \\ @V{f}VV @VV{g}V \\ X_1 @>{b}>> Y_1 \end{CD} $$

The arrow category carries the seven-term adjoint chain $$ \boxed{ \operatorname{coker} \dashv L_1 \dashv \operatorname{ev}_1 \dashv H \dashv \operatorname{ev}_0 \dashv L_0 \dashv \ker }. $$

Here

$$ \operatorname{ev}_i(X_0\xrightarrow{f}X_1)=X_i, \qquad L_1(A)=(0\to A), \qquad H(A)=(A\xrightarrow{\operatorname{id}_A}A), \qquad L_0(A)=(A\to 0). $$

1. The complete statement

For every arrow $f:X_0\to X_1$ and every object $A\in\mathcal A$, there are natural bijections

$$ \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A), $$ $$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_1f), $$ $$ \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_1f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA), $$ $$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_0f), $$ $$ \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_0f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A), $$

and

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$

2. The adjunction $\operatorname{coker}\dashv L_1$

We prove

$$ \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A). $$

Let $q_f:X_1\to\operatorname{coker}f$ be the cokernel morphism. A morphism $f\to L_1A$ is a commutative square

A morphism $f\to L_1A$.

$$ \begin{CD} X_0 @>{0}>> 0 \\ @V{f}VV @VV{0}V \\ X_1 @>{b}>> A \end{CD} $$

Since the top arrow is necessarily zero, commutativity is exactly the condition $bf=0$. By the universal property of the cokernel, this is equivalent to the existence of a unique morphism

$$ \overline b:\operatorname{coker}f\longrightarrow A $$

such that $b=\overline b\,q_f$. Hence

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A) \cong \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A). $$

So

$$ \boxed{\operatorname{coker}\dashv L_1}. $$

3. The adjunction $L_1\dashv\operatorname{ev}_1$

We prove

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_1). $$

A morphism $L_1A\to f$ is a commutative square

A morphism $L_1A\to f$.

$$ \begin{CD} 0 @>{0}>> X_0 \\ @V{0}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$

The commutativity condition $f\circ 0=b\circ 0$ is automatic, so the square is determined uniquely by $b:A\to X_1$. Therefore

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_1) = \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_1f). $$

Thus

$$ \boxed{L_1\dashv\operatorname{ev}_1}. $$

4. The adjunction $\operatorname{ev}_1\dashv H$

We prove

$$ \operatorname{Hom}_{\mathcal A}(X_1,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA). $$

A morphism $f\to HA$ is a commutative square

A morphism $f\to HA$.

$$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{\operatorname{id}_A}V \\ X_1 @>{b}>> A \end{CD} $$

Commutativity means $\operatorname{id}_A\circ a=bf$, hence $a=bf$. So once $b:X_1\to A$ is chosen, the top arrow is forced. Therefore

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA) \cong \operatorname{Hom}_{\mathcal A}(X_1,A). $$

Hence

$$ \boxed{\operatorname{ev}_1\dashv H}. $$

5. The adjunction $H\dashv\operatorname{ev}_0$

We prove

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_0). $$

A morphism $HA\to f$ is a commutative square

A morphism $HA\to f$.

$$ \begin{CD} A @>{a}>> X_0 \\ @V{\operatorname{id}_A}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$

Commutativity means $fa=b\,\operatorname{id}_A$, hence $b=fa$. So once $a:A\to X_0$ is chosen, the bottom arrow is forced. Therefore

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_0). $$

Hence

$$ \boxed{H\dashv\operatorname{ev}_0}. $$

6. The adjunction $\operatorname{ev}_0\dashv L_0$

We prove

$$ \operatorname{Hom}_{\mathcal A}(X_0,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A). $$

A morphism $f\to L_0A$ is a commutative square

A morphism $f\to L_0A$.

$$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{0}V \\ X_1 @>{0}>> 0 \end{CD} $$

The commutativity condition $0\circ a=0\circ f$ is automatic, so such a square is determined uniquely by $a:X_0\to A$. Therefore

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A) \cong \operatorname{Hom}_{\mathcal A}(X_0,A) = \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_0f,A). $$

Hence

$$ \boxed{\operatorname{ev}_0\dashv L_0}. $$

7. The adjunction $L_0\dashv\ker$

We prove

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$

Let $i_f:\ker f\to X_0$ be the kernel morphism. A morphism $L_0A\to f$ is a commutative square

A morphism $L_0A\to f$.

$$ \begin{CD} A @>{a}>> X_0 \\ @V{0}VV @VV{f}V \\ 0 @>{0}>> X_1 \end{CD} $$

Commutativity means $fa=0$. By the universal property of the kernel, this is equivalent to the existence of a unique morphism

$$ \overline a:A\longrightarrow\ker f $$

such that $a=i_f\overline a$. Hence

$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$

Therefore

$$ \boxed{L_0\dashv\ker}. $$

8. Summary of all six special squares

For convenience, here are the six special commuting squares that appear in the proof:

$$ \begin{CD} X_0 @>{0}>> 0 \\ @V{f}VV @VV{0}V \\ X_1 @>{b}>> A \end{CD} \qquad \begin{CD} 0 @>{0}>> X_0 \\ @V{0}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$ $$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{\operatorname{id}_A}V \\ X_1 @>{b}>> A \end{CD} \qquad \begin{CD} A @>{a}>> X_0 \\ @V{\operatorname{id}_A}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$ $$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{0}V \\ X_1 @>{0}>> 0 \end{CD} \qquad \begin{CD} A @>{a}>> X_0 \\ @V{0}VV @VV{f}V \\ 0 @>{0}>> X_1 \end{CD} $$

9. Conceptual explanation via Kan extensions

The middle five functors come from restriction along the endpoint inclusions

$$ i_0:\{0\}\hookrightarrow [1], \qquad i_1:\{1\}\hookrightarrow [1]. $$

Restriction gives

$$ i_0^*=\operatorname{ev}_0, \qquad i_1^*=\operatorname{ev}_1. $$

For $i_1$ one gets

$$ \operatorname{Lan}_{i_1}=L_1, \qquad \operatorname{Ran}_{i_1}=H, $$

hence

$$ L_1\dashv\operatorname{ev}_1\dashv H. $$

For $i_0$ one gets

$$ \operatorname{Lan}_{i_0}=H, \qquad \operatorname{Ran}_{i_0}=L_0, $$

hence

$$ H\dashv\operatorname{ev}_0\dashv L_0. $$

The overlap $H=\operatorname{Ran}_{i_1}=\operatorname{Lan}_{i_0}$ glues the two triples into the five-term chain

$$ L_1\dashv\operatorname{ev}_1\dashv H\dashv\operatorname{ev}_0\dashv L_0. $$

Finally, kernel and cokernel extend the chain at the ends:

$$ \operatorname{coker}\dashv L_1, \qquad L_0\dashv\ker. $$

So the seven-term chain is built from two overlapping Kan-extension triples, completed by cokernel and kernel.

This document uses MathJax with the amscd package so that the commutative squares render directly in the browser.