surjective homomorphism between unitary rings
Theorem. Let $f$ be a surjective^{} homomorphism^{} from a unitary ring $R$ to another unitary ring ${R}^{\prime}$. Then

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$f(1)={\mathrm{\hspace{0.33em}1}}^{\prime},$

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$f({a}^{1})={(f(a))}^{1}$ for all elements $a$ belonging to the group of units of $R$.
Proof. ${1}^{\circ}$. In a ring, the identity element is unique, whence it suffices to show that $f(1)$ has the properties required for the unity of the ring ${R}^{\prime}$. When ${a}^{\prime}$ is an arbitrary element of this ring, there is by the surjectivity an element $a$ of $R$ such that $f(a)={a}^{\prime}$. Thus we have
$$f(1){a}^{\prime}=f(1)f(a)=f(1a)=f(a)={a}^{\prime},{a}^{\prime}f(1)=f(a)f(1)=f(a1)=f(a)={a}^{\prime}.$$ 
${2}^{\circ}$. Let $a$ be a unit of $R$. Then
$$f(a)f({a}^{1})=f(a{a}^{1})=f(1)={\mathrm{\hspace{0.33em}1}}^{\prime},f({a}^{1})f(a)=f({a}^{1}a)=f(1)={\mathrm{\hspace{0.33em}1}}^{\prime},$$ 
whence $f({a}^{1})$ is a multiplicative inverse of $f(a)$.
Title  surjective homomorphism between unitary rings 

Canonical name  SurjectiveHomomorphismBetweenUnitaryRings 
Date of creation  20130322 19:10:22 
Last modified on  20130322 19:10:22 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  6 
Author  pahio (2872) 
Entry type  Theorem 
Classification  msc 16B99 
Classification  msc 13B10 
Related topic  IsomorphismSwappingZeroAndUnity 