determinant map of k1 of a ring

Let \(R\) be a commutative ring.
Then the determinant map of infinite general linear group

\begin{align*} \mathrm{det}(-): \mathrm{GL}(R) \rightarrow R^{\times} \end{align*}

induces an algebraic group-epimorphism

\begin{align*} K_1(R) \rightarrow R^{\times} \end{align*}