Morita Bicategories and Projective Generators

Zero and one are central: for every $a\in A$ distributivity gives $a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0$, and cancelling $a\cdot 0$ in the additive group yields $a\cdot 0=0$, while $0\cdot a=(0+0)\cdot a=0\cdot a+0\cdot a$ similarly yields $0\cdot a=0$; hence $0\cdot a=0=a\cdot 0$ and $0\in Z(A)$. Likewise $1\cdot a=a=a\cdot 1$ for every $a$ by the identity law, so $1\in Z(A)$.

The center is closed under addition: if $z,z^{\mathrm{\prime }}\in Z(A)$ and $a\in A$, then $(z+z^{\mathrm{\prime }})\cdot a=z\cdot a+z^{\mathrm{\prime }}\cdot a=a\cdot z+a\cdot z^{\mathrm{\prime }}=a\cdot (z+z^{\mathrm{\prime }})$ by the two distributive laws, so $z+z^{\mathrm{\prime }}\in Z(A)$.

The center is closed under multiplication: if $z,z^{\mathrm{\prime }}\in Z(A)$ and $a\in A$, then $(z{z}^{\mathrm{\prime }})\cdot a=z\cdot (z^{\mathrm{\prime }}\cdot a)=z\cdot (a\cdot z^{\mathrm{\prime }})=(z\cdot a)\cdot z^{\mathrm{\prime }}=(a\cdot z)\cdot z^{\mathrm{\prime }}=a\cdot (z\cdot z^{\mathrm{\prime }})$, using associativity of multiplication together with the centrality of $z$ and then of $z^{\mathrm{\prime }}$; hence $z{z}^{\mathrm{\prime }}\in Z(A)$.

The center is commutative: for $z,z^{\mathrm{\prime }}\in Z(A)$, centrality of $z$ evaluated at $a=z^{\mathrm{\prime }}$ gives $z{z}^{\mathrm{\prime }}=z^{\mathrm{\prime }}z$, so multiplication in $Z(A)$ is commutative.

The equivalence $A$ commutative $\iff$ $Z(A)=A$ holds: if $A$ is commutative then $za=az$ for all $z,a\in A$ by [F3], so every element of $A$ is central and $Z(A)=A$; conversely if $Z(A)=A$, then for arbitrary $x,y\in A$ the element $x$ lies in $Z(A)$ and hence $xy=yx$, so $A$ is commutative by [F3].

The center is closed under additive inverses: if $z\in Z(A)$ and $a\in A$, then $(-z)\cdot a+z\cdot a=(-z+z)\cdot a=0\cdot a=0$ by step 1.1, so $(-z)\cdot a$ is the additive inverse of $z\cdot a$ and therefore equals $-(z\cdot a)$ by uniqueness of additive inverses in $(A,+,0)$; symmetrically $a\cdot (-z)=-(a\cdot z)$. Since $z$ is central, $-(z\cdot a)=-(a\cdot z)$, so $(-z)\cdot a=a\cdot (-z)$ and $-z\in Z(A)$.

By steps 1.1, 1.2, 2.1 and 1.3 the subset $Z(A)$ contains $1$ and is closed under addition, additive inverses and multiplication, so it is a subring of $A$ by [F2]; in particular it is a ring in its own right, with the addition, multiplication, zero and identity inherited from $A$.

Steps 1.1-1.3 and 2.1 supply the closure conditions of the center with its identity, step 3.1 assembles them into the statement that $Z(A)$ is a subring of $A$, step 1.4 shows that this subring is commutative, and step 1.5 gives the asserted equivalence between commutativity of $A$ and $Z(A)=A$; every element considered lies in $A$ and no choice principle is used.

(Addition makes $E$ an abelian group.) The set $E=\mathcal{C}(P,P)$ is a hom-set of the preadditive category $\mathcal{C}$, hence an abelian group under its addition, with zero $0_{P,P}$ and additive inverses $-f$; this is axiom (R1) of [F2] for $E$.

(Composition is an associative unital operation on $E$.) If $f,g\in E$ then $g\circ f:P\to P$, so composition restricts to a binary operation on $E$; it is associative by [F3], and the identity morphism $1_P$ lies in $E$ and satisfies $1_P\circ f=f=f\circ 1_P$ by [F3]. Hence $(E,\circ ,1_P)$ is a monoid, which is axiom (R2).

(Both distributive laws and bilinearity.) For $f,g,h\in E$, bilinearity of composition in the preadditive category gives $h\circ (f+g)=h\circ f+h\circ g$ and $(f+g)\circ h=f\circ h+g\circ h$; these are the two distributive laws (R3) of [F2], and they say exactly that composition is bilinear in both variables on $E$.

($E$ is a unital ring.) By steps 1.1, 1.2 and 1.3 the set $E$ with addition and composition satisfies (R1), (R2) and (R3) of [F2], so $E$ is a unital ring whose identity is $1_P$; no element outside the given category is chosen.

(The reversed multiplication is the opposite ring.) Define $f\star g:=g\circ f$ on $E$; then $(E,+,\star ,1_P)$ is exactly the opposite ring of [F4] applied to the ring of step 2.1, because [F4] verifies the ring axioms for the reversed multiplication on the same abelian group with the same identity.

(Module case.) If $\mathcal{C}$ is the category of left $R$-modules, then $\mathcal{C}(P,P)={\mathrm{Hom}}_R(P,P)$ with pointwise addition and composition, so the ring constructed in step 2.1 is exactly the published endomorphism ring ${\mathrm{End}}_R(P)$ of [F5].

Steps 1.1-1.3 verify the ring axioms for ${\mathrm{End}}_{\mathcal{C}}(P)$ and give the bilinearity of composition, step 2.1 assembles them into the unital ring structure with identity $1_P$, step 3.1 identifies ${\mathrm{End}}_{\mathcal{C}}(P{)}^{\mathrm{o}\mathrm{p}}$, and step 3.2 matches the published module-case definition; nothing outside $\mathcal{C}$ is chosen and no choice principle is used.

($P^{\vee }$ is an $(A,B)$-bimodule.) For $a\in A$ and $\phi \in P^{\vee }$, the prescription $(a\phi )(p):=\phi (pa)$ is additive in $p$ because the right $A$-action and $\phi$ are additive, and it is left $B$-linear because $P$ is a bimodule and $\phi$ is left $B$-linear: $(a\phi )(bp)=\phi ((bp)a)=\phi (b(pa))=b\phi (pa)=b(a\phi )(p)$. For $b\in B$ the prescription $(\phi b)(p):=\phi (p)b$ is additive and left $B$-linear because $(\phi b)(b^{\mathrm{\prime }}p)=\phi (b^{\mathrm{\prime }}p)b=(b^{\mathrm{\prime }}\phi (p))b=b^{\mathrm{\prime }}(\phi (p)b)$. The module laws for $a\mapsto a\phi$ and $b\mapsto \phi b$ follow from the ring laws of $A$ and $B$ and the module laws of $P$, and the two actions commute, $(a\phi )b=a(\phi b)$, because both sides send $p$ to $\phi (pa)b$; hence $P^{\vee }$ is an $(A,B)$-bimodule.

(Existence of a dual basis.) Since $P$ is finitely generated, it has a finite generating set $x_1,\dots ,x_n$ by [F3]; the maps ${\beta }_i:B\to P$, $b\mapsto b{x}_i$, are left $B$-linear, so the universal property of the direct sum $B^n$ gives a unique left $B$-linear $q:B^n\to P$ with $q{\mathrm{ȷ}}_i={\beta }_i$. This $q$ is surjective: every $p\in P$ is a finite $B$-linear combination ${\sum }_i{r}_i{x}_i$ by [F3], and ${\sum }_i{r}_i{x}_i=q({\sum }_i{\mathrm{ȷ}}_i(r_i))$. By [F4] the epimorphism $q$ onto the projective module $P$ has a section $s:P\to B^n$ with $qs=1_P$; write ${\pi }_i:B^n\to B$ for the coordinate projections. Setting $p_i:=x_i$ and ${\phi }_i:={\pi }_i\circ s$ gives left $B$-linear maps ${\phi }_i:P\to B$, and for every $p$ the description of elements of a direct sum [F5] gives $s(p)={\sum }_i{\mathrm{ȷ}}_i({\pi }_i(s(p)))$, so $p=qs(p)={\sum }_i{\pi }_i(s(p))q{\mathrm{ȷ}}_i(1)={\sum }_i{\phi }_i(p)p_i$. Only finitely many objects are selected inside the given finite generating set, so no choice principle is used.

(The evaluation pairing is balanced.) For fixed $y\in Y$ the map $\phi \mapsto (p\mapsto \phi (p)y)$ is additive by [F2], and for fixed $\phi$ the map $y\mapsto (p\mapsto \phi (p)y)$ is additive; the pairing $(\phi ,y)\mapsto (p\mapsto \phi (p)y)$ from $P^{\vee }\times Y$ to ${\mathrm{Hom}}_B(P,Y)$ is therefore additive in each variable. It is $B$-balanced because $(\phi b)(p)y=\phi (p)by=\phi (p)(by)$ for $b\in B$: forming $(\phi b,y)$ and $(\phi ,by)$ gives the same map $P\to Y$.

(The evaluation map exists and is natural.) By the universal property of the tensor product [F6], the balanced pairing of step 1.3 induces a unique group homomorphism $\mathrm{ev}:P^{\vee }{\otimes }_{B}Y\to {\mathrm{Hom}}_B(P,Y)$ with $\mathrm{ev}(\phi \otimes y)=(p\mapsto \phi (p)y)$. For a left $B$-module map $g:Y\to Y^{\mathrm{\prime }}$, the composites ${\mathrm{ev}}_{Y^{\mathrm{\prime }}}\circ (1\otimes g)$ and $g_{\ast }\circ {\mathrm{ev}}_Y$ are group homomorphisms that agree on every elementary tensor $\phi \otimes y$, both sending it to $(p\mapsto \phi (p)g(y))$ by [F8] and [F2]; since the elementary tensors generate the tensor product additively, the two maps are equal, which is naturality in $Y$.

(A candidate inverse from the dual basis.) Fix the dual basis $p_i,{\phi }_i$ of step 1.2 and define $\delta :{\mathrm{Hom}}_B(P,Y)\to P^{\vee }{\otimes }_{B}Y$ by $\delta (h):={\sum }_i{\phi }_i\otimes h(p_i)$; this is a finite sum, and it is additive in $h$ because each evaluation $h\mapsto h(p_i)$ and the tensor product are additive.

(The two composites are identities.) For $h\in {\mathrm{Hom}}_B(P,Y)$, applying $\mathrm{ev}$ to $\delta (h)$ gives the map $p\mapsto {\sum }_i{\phi }_i(p)h(p_i)$, which equals $h(p)$ because $h$ is left $B$-linear and $p={\sum }_i{\phi }_i(p)p_i$; hence $\mathrm{ev}\circ \delta =1$. For an elementary tensor, $\delta (\mathrm{ev}(\phi \otimes y))={\sum }_i{\phi }_i\otimes \phi (p_i)y$. The element ${\sum }_i{\phi }_i\phi (p_i)$ of $P^{\vee }$ equals $\phi$, since for every $p$ one has $({\sum }_i{\phi }_i\phi (p_i))(p)={\sum }_i{\phi }_i(p)\phi (p_i)=\phi ({\sum }_i{\phi }_i(p)p_i)=\phi (p)$ by left $B$-linearity of $\phi$ and the dual-basis formula; hence ${\sum }_i{\phi }_i\otimes \phi (p_i)y=({\sum }_i{\phi }_i\phi (p_i))\otimes y=\phi \otimes y$, so $\delta \circ \mathrm{ev}$ is the identity on elementary tensors and therefore on all of $P^{\vee }{\otimes }_{B}Y$. Thus $\mathrm{ev}$ is a bijection with inverse $\delta$, and it is in particular an isomorphism of abelian groups.

($A$-linearity.) On $P^{\vee }{\otimes }_{B}Y$ the left $A$-action $a(\phi \otimes y)=(a\phi )\otimes y$ is the one of [F7] for the $(A,B)$-bimodule $P^{\vee }$ of step 1.1, and on ${\mathrm{Hom}}_B(P,Y)$ we use $(a\psi )(p)=\psi (pa)$, which is again a left $B$-linear map by the same bimodule computation as in step 1.1. For $a\in A$ and $\phi \otimes y$, $\mathrm{ev}(a(\phi \otimes y))$ sends $p$ to $(a\phi )(p)y=\phi (pa)y$, and $a\mathrm{ev}(\phi \otimes y)$ sends $p$ to $\mathrm{ev}(\phi \otimes y)(pa)=\phi (pa)y$; the two sides agree on elementary tensors and hence everywhere, so $\mathrm{ev}$ is left $A$-linear.

Steps 1.1-1.3 and 2.1 construct the $(A,B)$-bimodule $P^{\vee }$ and the natural evaluation map, steps 2.2 and 3.1 show it is an isomorphism with the stated inverse built from any dual basis, and step 3.2 upgrades it to a left $A$-module isomorphism. Retaining only the group structures, steps 2.1 and 3.1 give the natural isomorphism ${\mathrm{Hom}}_B(P,-)\cong {\mathrm{Hom}}_B(P,B){\otimes }_B-$ of the final sentence: neither the existence of the dual basis nor the two composite computations uses the $A$-action, so for a left $B$-module $P$ finitely generated and projective the result applies with any right $A$-structure on $P$ or with none. The only selections made lie inside the finite data of a generating set and a section, so no choice principle is used.

($H$ is an additive functor to $A\text{-Mod}$.) By [F2] the ring $A=E^{\mathrm{o}\mathrm{p}}$ is unital with identity $1_P^{\mathrm{o}\mathrm{p}}$ and composition in $E$ is bilinear. For $Y\in \mathcal{C}$, $H(Y)=\mathcal{C}(P,Y)$ is an abelian group by [F3], and the prescription $e^{\mathrm{o}\mathrm{p}}\cdot h:=h\circ e$ makes it a left $A$-module: $((e+f{)}^{\mathrm{o}\mathrm{p}})\cdot h=h\circ (e+f)=h\circ e+h\circ f$ and $(e^{\mathrm{o}\mathrm{p}}+f^{\mathrm{o}\mathrm{p}})\cdot h$ agree, $(e^{\mathrm{o}\mathrm{p}}{f}^{\mathrm{o}\mathrm{p}})\cdot h=((fe{)}^{\mathrm{o}\mathrm{p}})\cdot h=h\circ (fe)=(h\circ f)\circ e=e^{\mathrm{o}\mathrm{p}}\cdot (f^{\mathrm{o}\mathrm{p}}\cdot h)$, and $1_P^{\mathrm{o}\mathrm{p}}\cdot h=h\circ 1_P=h$. For $u:X\to Y$, postcomposition $u_{\ast }=H(u)$ is additive by [F3] and $A$-linear: $u_{\ast }(e^{\mathrm{o}\mathrm{p}}\cdot h)=u\circ h\circ e=e^{\mathrm{o}\mathrm{p}}\cdot u_{\ast }(h)$. Identities and composites are preserved because postcomposition is associative and unital, so $H$ is a functor $\mathcal{C}\to A\text{-Mod}$ whose hom-maps are group homomorphisms, that is, an additive functor by [F10].

($H$ is faithful.) By [F6] the functor $\mathcal{C}(P,-)$ is faithful: for all $u\mathrm{\ne }v:X\to Y$ there is $h:P\to X$ with $u\circ h\mathrm{\ne }v\circ h$. The underlying functions of $H(u)$ and $\mathcal{C}(P,-)(u)$ coincide, so $H(u)\mathrm{\ne }H(v)$, and $H$ is faithful in the stated elementwise form.

(Nonzero objects are detected.) Suppose $H(Z)=0$, so $H(Z)$ is the zero group and every morphism $P\to Z$ is zero. By [F6] the canonical morphism $f:{\coprod }_{u\in \mathcal{C}(P,Z)}P\to Z$ is an epimorphism; its index set is the singleton $\{0_{P,Z}\}$, so by [F7] the coproduct is $P$ and $f$ is the zero morphism $0:P\to Z$. The image of $0:P\to Z$ is the zero subobject, the same image as that of the zero morphism $0\to Z$ out of the zero object, so $\mathrm{coker}f\cong \mathrm{coker}(0\to Z)\cong Z$ by [F8]. Since $f$ is epic, [F9] gives $\mathrm{coker}f=0$; hence $Z\cong 0$, that is, $Z=0$.

($H$ preserves coproducts.) By clause (iii) of [F1], the functor $\mathcal{C}(P,-)$ preserves every set-indexed coproduct, and the additional $A$-module structure of step 1.1 only adds structure to the values: the canonical comparison maps for $H$ have the same underlying group homomorphisms as those of $\mathcal{C}(P,-)$, which are isomorphisms. Hence $H$ preserves every set-indexed coproduct.

($H$ carries short exact sequences to short exact sequences.) Let $0\to K\to E^{\mathrm{\prime }}\to M\to 0$ be short exact in $\mathcal{C}$. By [F4] the induced sequence $0\to H(K)\to H(E^{\mathrm{\prime }})\to H(M)\to 0$ is exact; the maps are $A$-linear by step 1.1, and $H$ is additive by step 1.1.

($H$ preserves kernels and cokernels.) Let $g:X\to Y$ be a morphism of $\mathcal{C}$. Applying step 2.2 to $0\to \ker g\to X\to \mathrm{im}g\to 0$ and using left exactness [F5] identifies $H(\ker g)$ with $\ker H(g)$; applying step 2.2 to $0\to \mathrm{im}g\to Y\to \mathrm{coker}g\to 0$ and using that $H(X)\to H(\mathrm{im}g)$ is surjective with kernel $H(\ker g)$ identifies $H(\mathrm{coker}g)$ with $\mathrm{coker}H(g)$. Hence $H$ preserves kernels and cokernels.

($H$ is exact.) By step 1.1, $H$ is additive, by [F5] it is left exact, and step 3.1 with [F11] makes it right exact as well: an additive functor preserving kernels and cokernels is exact. By the definitions of [F10] this means precisely that $H$ preserves every finite limit and every finite colimit that exists in $\mathcal{C}$, in addition to the kernels and cokernels just exhibited.

Steps 1.1-1.3 establish that $H$ is an additive functor and is faithful, and show that $H(Z)=0$ forces $Z=0$; step 2.1 gives coproduct preservation, and steps 2.2, 3.1 and 4.1 give exactness with preservation of kernels, cokernels and finite limits and colimits. Every construction uses only the given category, the object $P$ and postcomposition, so no element is chosen and no choice principle is used.

(Finitely generated implies coproduct preservation.) Assume $P$ is finitely generated, with finite generating set $S=\{x_1,\dots ,x_n\}$, so every element of $P$ is a finite $B$-linear combination of the $x_j$ by [F3]. Let $Y={⨁}_i{Y}_i$ be a set-indexed direct sum in $B$-Mod and let $f:P\to Y$. Each $f(x_j)$ has finite support by [F4], so the union $F$ of the finitely many supports is finite and $f(x_j)\in {⨁}_{i\in F}{Y}_i$ for every $j$; since the $x_j$ generate $P$ and the submodule ${⨁}_{i\in F}{Y}_i$ contains all their images, $f$ factors through the inclusion ${\mathrm{ȷ}}_F:{⨁}_{i\in F}{Y}_i\to Y$. The canonical comparison $c:{⨁}_i{\mathrm{Hom}}_B(P,Y_i)\to {\mathrm{Hom}}_B(P,Y)$, $(u_i)\mapsto {\sum }_i{\mathrm{ȷ}}_i\circ u_i$, has components ${\pi }_i\circ c(u)=u_i$ by [F4] and [F10], so $c$ is injective; and for arbitrary $f$ the finite-support family $({\pi }_i\circ f{)}_i$ satisfies $c(({\pi }_i\circ f{)}_i)={\sum }_i{\mathrm{ȷ}}_i\circ {\pi }_i\circ f=f$ by [F4], so $c$ is surjective. Hence ${\mathrm{Hom}}_B(P,-)$ preserves this coproduct, and projectivity of $P$ was not used.

(Coproduct preservation and projectivity imply finite generation.) Assume $P$ is projective and ${\mathrm{Hom}}_B(P,-)$ preserves every set-indexed coproduct. Let $\epsilon :B^{(P)}\to P$ be the canonical surjection of [F6] from the free module on the underlying set of $P$, and let $s:P\to B^{(P)}$ be a section of $\epsilon$, which exists because $P$ is projective: lift the identity of $P$ through the epimorphism by [F2]. Smallness identifies ${\mathrm{Hom}}_B(P,B^{(P)})$ with ${⨁}_{p\in P}{\mathrm{Hom}}_B(P,B)$ under the comparison, so the family ${\phi }_p:={\pi }_p\circ s$ has finite support: there is a finite subset $F\subseteq P$ with ${\phi }_p=0$ for $p\mathrm{\notin }F$. For every $x\in P$ the element $s(x)$ has support contained in $F$ by [F4], so $x=\epsilon (s(x))={\sum }_{p\in F}{\phi }_p(x)p$ is a finite $B$-linear combination of the finitely many elements $p\in F$; by [F3] the module $P$ is generated by $F$, hence finitely generated. No infinite choice is used: the section $s$ is a single existential instance and the set $F$ is computed from it.