Let $A$ ba a (not necessarily commutative) unital ring. The $K_1$ group of $A$ is defined as $K_1(A)=\pi_1((\mathrm{Proj}(A)^\simeq)^\mathrm{gp})$. Here $\mathrm{Proj}^\simeq$ is the core of the category of finitely generated projective left $A$-modules, which is a symmetric monoidal groupoid and is sent to a Picard groupoid by the group completion functor $(-)^\mathrm{gp}$, and $\pi_1$ means taking the automorphism group of an object. A ring map $A\to B$ induces a morphism $K_1(A)\to K_1(B)$.
I am considering the following problem
If $I\subset A$ is a nilpotent two-sided ideal, then $K_1(A)\to K_1(A/I)$ is surjective.
I can show that $\mathrm{Proj}(A)^\simeq\to\mathrm{Proj}(A/I)^\simeq$ is bijective on objects and surjective on their corresponding isomorphism groups. Also, the group completion $(-)^\mathrm{gp}$ is a left adjoint. Can I conclude from these facts that $K_1(A)\twoheadrightarrow K_1(A/I)$? I am not familiar with $2$-categorical stuffs, but these seem to me like a left-adjoint-preserves-epimorphism argument.
ps. The proof I know is more concrete. Note that there is an identification $K_1(A)\simeq GL(A)/E(A)$, where $GL(A)$ and $E(A)$ are respectively the colimits of $GL_n(A)$ (the group of $n$-by-$n$ invertible matrices) and $E_n(A)$ (the subgroup generated by $n$-by-$n$ elementary matrices, i.e. matrices with $1$'s on the diagonal and with one possible non-zero entry except the diagonal). Then the surjectivity follows easily from that we can lift an invertible matrix in $GL_n(A/I)$ to one in $GL_n(A)$.