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I'm wondering if there are homomorphisms $f$ between unitary rings $R,S$, such that $f(r)$ is invertible but it's inverse doesn't equal $f(r^{-1})$. That is only possible if the inverse isn't in the range of $f$. Nonetheless I couldn't find example of such rings and homomorphisms.

(Notice that of course we know that if $r$ is invertible, we have $f(r)^{-1}= f(r^{-1})$, but I only assume that $f(r)$ is invertible)

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    @Micah If $f(r)$ is invertible *and* $r$ is invertible, then the inverse of $f(r)$ is automatically $f(r)^{-1}$, since $rr^{-1}=1$ implies $f(r)f(r^{-1})=1$. But what if $f(r)$ is invertible but $r$ isn't ?2012-12-02

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No. Let $r\in R$, and suppose $r^{-1}\in R$ exists. Then $f(rr^{-1})=f(1_R)=1_S=f(r)f(r^{-1})$, so $f(r^{-1})$ is an inverse for $f(r)$, so by uniqueness of inverses in $S$ we have $f(r^{-1})=f(r)^{-1}$

If $r^{-1}$ does not exist, then your question is ill-posed. However, there are many homomorphisms (such as $R=\mathbb{Z}\to S=\mathbb{Q}$) such that the image $f(r)$ of some elements is invertible, but the original $r$ is not.