Editing Homomorphism (section)

=== Isomorphism ===
An [[isomorphism]] between [[algebraic structure]]s of the same type is commonly defined as a [[bijective]] homomorphism.<ref name="Birkhoff.1967">{{cite book | last1=Birkhoff | first1=Garrett | title=Lattice theory | orig-year=1940 | publisher=[[American Mathematical Society]] | location=Providence, Rhode Island | edition=3rd | series=American Mathematical Society Colloquium Publications | isbn=978-0-8218-1025-5 | mr=598630 | year=1967 | volume=25}}</ref>{{rp|134}}<ref name="Burris.Sankappanavar.2012">{{cite book | url=http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra2012.pdf | isbn=978-0-9880552-0-9 | first1=Stanley N. |last1=Burris | first2=H. P. |last2=Sankappanavar | title=A Course in Universal Algebra | year=2012 | publisher=S. Burris and H.P. Sankappanavar }}</ref>{{rp|28}}

In the more general context of [[category theory]], an isomorphism is defined as a [[morphism]] that has an [[inverse function|inverse]] that is also a morphism. In the specific case of algebraic structures, the two definitions are equivalent, although they may differ for non-algebraic structures, which have an underlying set.

More precisely, if 

<math display="block">f: A\to B</math>

is a (homo)morphism, it has an inverse if there exists a homomorphism

<math display="block">g: B\to A</math>

such that

<math display="block">f\circ g = \operatorname{Id}_B \qquad \text{and} \qquad g\circ f = \operatorname{Id}_A.</math>

If <math>A</math> and <math>B</math> have underlying sets, and <math>f: A \to B</math> has an inverse <math>g</math>, then <math>f</math> is bijective. In fact, <math>f</math> is [[injective]], as <math>f(x) = f(y)</math> implies <math>x = g(f(x)) = g(f(y)) = y</math>, and <math>f</math> is [[surjective]], as, for any <math>x</math> in <math>B</math>, one has <math>x = f(g(x))</math>, and <math>x</math> is the image of an element of <math>A</math>.

Conversely, if <math>f: A \to B</math> is a bijective homomorphism between algebraic structures, let <math>g: B \to A</math> be the map such that <math>g(y)</math> is the unique element <math>x</math> of <math>A</math> such that <math>f(x) = y</math>. One has <math>f \circ g = \operatorname{Id}_B \text{ and } g \circ f = \operatorname{Id}_A,</math> and it remains only to show that {{math|''g''}} is a homomorphism. If <math>*</math> is a binary operation of the structure, for every pair <math>x</math>, <math>y</math> of elements of <math>B</math>, one has

<math display="block">g(x*_B y) = g(f(g(x))*_Bf(g(y))) = g(f(g(x)*_A g(y))) = g(x)*_A g(y),</math>

and <math>g</math> is thus compatible with <math>*.</math> As the proof is similar for any [[arity]], this shows that <math>g</math> is a homomorphism.

This proof does not work for non-algebraic structures. For example, for [[topological space]]s, a morphism is a [[continuous map]], and the inverse of a bijective continuous map is not necessarily continuous. An isomorphism of topological spaces, called [[homeomorphism]] or [[bicontinuous function|bicontinuous map]], is thus a bijective continuous map, whose inverse is also continuous.