[proofplan] We first use the [normality of the stable elementary subgroup](/theorems/8651) to ensure that $GL(R)/E(R)$ is a group. Then the [Whitehead lemma](/theorems/8652) shows that every commutator in $GL(R)$ lies in $E(R)$, so the quotient is abelian. Finally, a unital ring homomorphism acts entrywise on invertible matrices, respects stabilization, and sends elementary matrices to elementary matrices; therefore it descends to the quotient, with identity and composition inherited from entrywise application. [/proofplan]

[step:Use normality of the stable elementary subgroup to form the quotient] Let $R$ be a unital ring. By [citetheorem:8651], the subgroup $E(R)$ is normal in $GL(R)$. Hence the set of left cosets \begin{align*} GL(R)/E(R) \end{align*} has the [quotient group](/theorems/790) structure whose product is given by \begin{align*} (gE(R))(hE(R))=(gh)E(R) \end{align*} for $g,h\in GL(R)$. This product is well-defined precisely because $E(R)\trianglelefteq GL(R)$. [/step]

[step:Show every commutator vanishes in the quotient]Let $q_R:GL(R)\to GL(R)/E(R)$ denote the quotient homomorphism. By the Whitehead lemma, stated as [citetheorem:8652], the stable elementary subgroup satisfies \begin{align*} E(R)=[GL(R),GL(R)]. \end{align*} In particular, for every pair $g,h\in GL(R)$, the commutator \begin{align*} [g,h]=ghg^{-1}h^{-1} \end{align*} belongs to $E(R)=\ker q_R$. Therefore \begin{align*} q_R(g)q_R(h)q_R(g)^{-1}q_R(h)^{-1}=q_R([g,h])=E(R), \end{align*} so $q_R(g)q_R(h)=q_R(h)q_R(g)$. Since every element of $GL(R)/E(R)$ has the form $q_R(g)$ for some $g\in GL(R)$, the quotient group is abelian.[/step]

[guided]Let $q_R:GL(R)\to GL(R)/E(R)$ be the canonical quotient homomorphism, so $q_R(g)=gE(R)$ for $g\in GL(R)$. To prove that the quotient is abelian, it is enough to prove that the commutator of any two quotient elements is the identity coset. Take arbitrary elements $g,h\in GL(R)$. Their group commutator is \begin{align*} [g,h]=ghg^{-1}h^{-1}. \end{align*} The Whitehead lemma, [citetheorem:8652], says that the stable elementary subgroup contains exactly the commutator subgroup of the stable general linear group: \begin{align*} E(R)=[GL(R),GL(R)]. \end{align*} Thus $[g,h]\in E(R)$. Since $E(R)$ is the kernel of $q_R$, this gives \begin{align*} q_R([g,h])=E(R). \end{align*} Now use that $q_R$ is a [group homomorphism](/page/Group%20Homomorphism). We compute \begin{align*} q_R(g)q_R(h)q_R(g)^{-1}q_R(h)^{-1}=q_R(ghg^{-1}h^{-1})=q_R([g,h])=E(R). \end{align*} This says that the commutator of $q_R(g)$ and $q_R(h)$ is the identity element of the quotient group. Therefore $q_R(g)q_R(h)=q_R(h)q_R(g)$. Because $g$ and $h$ were arbitrary and every element of the quotient is represented by some element of $GL(R)$, the group $GL(R)/E(R)$ is abelian.[/guided]

[step:Define the map induced by a unital ring homomorphism] Let $\varphi:R\to S$ be a unital ring homomorphism. For each integer $n\ge 1$, define \begin{align*} \varphi_n:GL_n(R)\to GL_n(S) \end{align*} by applying $\varphi$ to every matrix entry. If $A\in GL_n(R)$, then $\varphi_n(A)$ is invertible with inverse $\varphi_n(A^{-1})$, because $\varphi_n(AA^{-1})=I_n$ and $\varphi_n(A^{-1}A)=I_n$. The maps $\varphi_n$ commute with block-sum stabilization since entrywise application sends \begin{align*} \begin{pmatrix} A & 0 \end{align*} \begin{align*} 0 & 1 \end{pmatrix} \end{align*} to \begin{align*} \begin{pmatrix} \varphi_n(A) & 0 \end{align*} \begin{align*} 0 & 1 \end{pmatrix}. \end{align*} Therefore the maps $\varphi_n$ induce a group homomorphism \begin{align*} \varphi_{GL}:GL(R)\to GL(S). \end{align*} [/step]

[step:Descend the entrywise map through the elementary subgroups] For distinct indices $i\ne j$ and an element $r\in R$, let $e_{ij}(r)$ denote the corresponding elementary matrix in the stable group $GL(R)$. Entrywise application gives \begin{align*} \varphi_{GL}(e_{ij}(r))=e_{ij}(\varphi(r)). \end{align*} Hence $\varphi_{GL}$ sends the generators of $E(R)$ into $E(S)$, and therefore \begin{align*} \varphi_{GL}(E(R))\subseteq E(S). \end{align*} Consequently, the rule \begin{align*} \varphi_*(gE(R))=\varphi_{GL}(g)E(S) \end{align*} defines a well-defined group homomorphism \begin{align*} \varphi_*:GL(R)/E(R)\to GL(S)/E(S). \end{align*} Indeed, if $gE(R)=hE(R)$, then $h^{-1}g\in E(R)$, so \begin{align*} \varphi_{GL}(h)^{-1}\varphi_{GL}(g)=\varphi_{GL}(h^{-1}g)\in E(S), \end{align*} and hence $\varphi_{GL}(g)E(S)=\varphi_{GL}(h)E(S)$. [/step]

[step:Verify identity and composition] For the identity homomorphism $\operatorname{id}_R:R\to R$, entrywise application is the identity on every $GL_n(R)$ and hence on $GL(R)$. Therefore \begin{align*} (\operatorname{id}_R)_*=\operatorname{id}_{GL(R)/E(R)}. \end{align*} Now let $R\xrightarrow{\varphi}S\xrightarrow{\psi}T$ be unital ring homomorphisms. For every stable matrix representative $g\in GL(R)$, entrywise application satisfies \begin{align*} (\psi\circ\varphi)_{GL}(g)=\psi_{GL}(\varphi_{GL}(g)). \end{align*} Passing to quotient cosets gives \begin{align*} (\psi\circ\varphi)_*(gE(R))=\psi_*(\varphi_*(gE(R))). \end{align*} Since every element of $GL(R)/E(R)$ is represented by such a coset, we have \begin{align*} (\psi\circ\varphi)_*=\psi_*\circ\varphi_*. \end{align*} Thus the abelian quotient $GL(R)/E(R)$ is functorial in $R$ with respect to unital ring homomorphisms. [/step]