first isomorphism theorem


Let Σ be a fixed signaturePlanetmathPlanetmathPlanetmath, and 𝔄 and 𝔅 structuresMathworldPlanetmath for Σ. If f:𝔄𝔅 is a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, then there is a unique bimorphismPlanetmathPlanetmath ϕ:𝔄/ker(f)im(f) such that for all a𝔄, ϕ([[a]])=f(a). Furthermore, if f has the additional property that for each n and each n-ary relation symbol R of Σ,

R𝔅(f(a1),,f(an))ai[f(ai)=f(ai)R𝔄(a1,,an)],

then ϕ is an isomorphismMathworldPlanetmathPlanetmath.

Proof.

Since the homomorphic image of a Σ-structure is also a Σ-structure, we may assume that im(f)=𝔅.

Let =ker(f). Define a bimorphism ϕ:𝔄/𝔅:[[a]]f(a). To verify that ϕ is well defined, let aa. Then ϕ([[a]])=f(a)=f(a)=ϕ([[a]]). To show that ϕ is injectivePlanetmathPlanetmath, suppose ϕ([[a]])=ϕ([[a]]). Then f(a)=f(a), so aa. Hence [[a]]=[[a]]. To show that ϕ is a homomorphism, observe that for any constant symbol c of Σ we have ϕ([[c𝔄]])=f(c𝔄)=c𝔅. For each n and each n-ary function symbol F of Σ,

ϕ(F𝔄/([[a1]],,[[an]])) =ϕ([[F𝔄(a1,,an)]])
=f(F𝔄(a1,,an))
=F𝔅(f(a1),,f(an))
=F𝔅(ϕ([[a1]],,ϕ([[an]])).

For each n and each n-ary relation symbol R of Σ,

R𝔄/([[a1]],,[[an]]) R𝔄(a1,,an)
R𝔅(f(a1),,f(an))
R𝔅(ϕ([[a1]],,ϕ([[an]])).

Thus ϕ is a bimorphism.

Now suppose f has the additional property mentioned in the statement of the theorem. Then

R𝔅(ϕ([[a1]]),,ϕ([[an]])) R𝔅(f(a1),,f(an))
ai[aiaiR𝔄(a1,,an)]
R𝔄/([[a1]],,[[an]]).

Thus ϕ is an isomorphism. ∎

Title first isomorphism theoremPlanetmathPlanetmath
Canonical name FirstIsomorphismTheorem1
Date of creation 2013-03-22 13:50:42
Last modified on 2013-03-22 13:50:42
Owner almann (2526)
Last modified by almann (2526)
Numerical id 10
Author almann (2526)
Entry type Theorem
Classification msc 03C07