surjective homomorphism between unitary rings

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

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

• $f(a^{-1})={(f(a))}^{-1}$  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 $f(a)$.