Editing Bimodule (section)

== Further notions and facts ==
If ''M'' and ''N'' are ''R''-''S''-bimodules, then a map {{nowrap|''f'' : ''M'' → ''N''}} is a ''bimodule homomorphism'' if it is both a homomorphism of left ''R''-modules and of right ''S''-modules.

An ''R''-''S''-bimodule is actually the same thing as a left module over the ring {{nowrap|''R'' ⊗<sub>'''Z'''</sub> ''S''<sup>op</sup>}}, where ''S''<sup>op</sup> is the [[opposite ring]] of ''S'' (where the multiplication is defined with the arguments exchanged). Bimodule homomorphisms are the same as homomorphisms of left {{nowrap|''R'' ⊗<sub>'''Z'''</sub> ''S''<sup>op</sup>}} modules. Using these facts, many definitions and statements about modules can be immediately translated into definitions and statements about bimodules. For example, the [[category (mathematics)|category]] of all {{nowrap|''R''-''S''-bimodules}} is [[abelian category|abelian]], and the standard [[isomorphism theorems]] are valid for bimodules.
 
There are however some new effects in the world of bimodules, especially when it comes to the [[tensor product of modules|tensor product]]: if ''M'' is an {{nowrap|''R''-''S''-bimodule}} and ''N'' is an {{nowrap|''S''-''T''-bimodule}}, then the tensor product of ''M'' and ''N'' (taken over the ring ''S'') is an {{nowrap|''R''-''T''-bimodule}} in a natural fashion. This tensor product of bimodules is [[associative]] ([[up to]] a unique canonical [[isomorphism]]), and one can hence construct a category whose objects are the rings and whose morphisms are the bimodules. This is in fact a [[2-category]], in a canonical way – 2 morphisms between {{nowrap|''R''-''S''-bimodules}} ''M'' and ''N'' are exactly bimodule homomorphisms, i.e. functions
: <math>f: M \rightarrow N</math>
that satisfy
# <math>f(m+m') = f(m)+ f(m')</math>  
# <math>f(r.m.s) = r.f(m).s</math>,
for {{nowrap|''m'' ∈ ''M''}}, {{nowrap|''r'' ∈ ''R''}}, and {{nowrap|''s'' ∈ ''S''}}. One immediately verifies the interchange law for bimodule homomorphisms, i.e.
: <math>(f'\otimes g')\circ (f\otimes g) = (f'\circ f)\otimes(g'\circ g) </math>
holds whenever either (and hence the other) side of the equation is defined, and where <math>\circ</math> is the usual composition of homomorphisms. In this interpretation, the category {{nowrap|1='''End'''(''R'') = '''Bimod'''(''R'', ''R'')}} is exactly the [[monoidal category]] of {{nowrap|''R''-''R''-bimodules}} with the usual [[tensor product]] over ''R'' the tensor product of the category. In particular, if ''R'' is a [[commutative ring]], every left or right ''R''-module is canonically an {{nowrap|''R''-''R''-bimodule}}, which gives a monoidal embedding of the category {{nowrap|1= ''R''-'''Mod'''}} into {{nowrap|1='''Bimod'''(''R'', ''R'')}}. The case that ''R'' is a [[field (mathematics)|field]] ''K'' is a motivating example of a symmetric monoidal category, in which case {{nowrap|1=''R''-'''Mod''' = ''K''-'''Vect'''}}, the [[category of vector spaces]] over ''K'', with the usual tensor product {{nowrap|1=⊗ = ⊗<sub>''K''</sub>}} giving the monoidal structure, and with unit ''K''. We also see that a [[monoid (category theory)|monoid]] in {{nowrap|'''Bimod'''(''R'', ''R'')}} is exactly an ''R''-algebra.{{clarify|reason=Are we still requiring that ''R'' is commutative?|date=June 2024}}<ref name=arXiv>{{cite arXiv|last1=Street|first1=Ross|title=Categorical and combinatorial aspects of descent theory|date=20 Mar 2003|eprint=math/0303175}}</ref>
Furthermore, if ''M'' is an {{nowrap|''R''-''S''-bimodule}} and ''L'' is an {{nowrap|''T''-''S''-bimodule}}, then the [[set (mathematics)|set]] {{nowrap|Hom<sub>''S''</sub>(''M'', ''L'')}} of all ''S''-module homomorphisms from ''M'' to ''L'' becomes a {{nowrap|''T''-''R''-bimodule}} in a natural fashion. These statements extend to the [[derived functor]]s [[Ext functor|Ext]] and [[Tor functor|Tor]].

[[Profunctor]]s can be seen as a categorical generalization of bimodules.

Note that bimodules are not at all related to [[bialgebra]]s.