[proofplan] We prove the corrected statement: congruence matrices modulo relative elementary matrices describe the kernel of the reduction map $K_1(R)\to K_1(R/I)$, equivalently the image of relative $K_1$ in $K_1(R)$ in the long exact sequence of the homotopy fibre. The plus construction identifies $K_1(R)$ with $GL(R)/E(R)$ and $K_1(R/I)$ with $GL(R/I)/E(R/I)$. A class in the kernel is represented by a stable matrix whose reduction is elementary; lifting that elementary correction moves the representative into $GL(R,I)$. The remaining ambiguity is exactly multiplication by $E(R)\cap GL(R,I)=E(R,I)$. [/proofplan]

[step:Fix the stable groups and the relative elementary subgroup]Let \begin{align*} \alpha_R:GL(R)\to GL(R/I) \end{align*} denote the stable [group homomorphism](/page/Group%20Homomorphism) induced entrywise by the quotient map $R\to R/I$. By definition, \begin{align*} GL(R,I)=\ker(\alpha_R). \end{align*} Let \begin{align*} \epsilon_R:E(R)\to E(R/I) \end{align*} be the restriction of $\alpha_R$ to stable elementary subgroups, and define \begin{align*} E(R,I)=\ker(\epsilon_R)=E(R)\cap GL(R,I). \end{align*} The subgroup $E(R)$ is normal in $GL(R)$ by [[Stable Elementary Subgroup Is Normal](/theorems/8655)][citetheorem:8655]. Moreover $E(R)$ is perfect by [[Whitehead Lemma](/theorems/8652)][citetheorem:8652], so the plus construction of $BGL(R)$ with respect to $E(R)$ is defined. The same statements apply to $R/I$.[/step]

[guided]The theorem is a stable statement, so all groups in the proof are direct limits over matrix size. The map \begin{align*} \alpha_R:GL(R)\to GL(R/I) \end{align*} is obtained by applying the quotient homomorphism $R\to R/I$ to every matrix entry, after passing to the stable direct limit. Its kernel is the congruence subgroup \begin{align*} GL(R,I)=\ker(\alpha_R), \end{align*} that is, the subgroup represented by invertible matrices congruent to the identity matrix modulo $I$ after stabilization. We also need the elementary subgroup killed by the plus construction. Let \begin{align*} \epsilon_R:E(R)\to E(R/I) \end{align*} be the restriction of $\alpha_R$. We define \begin{align*} E(R,I)=\ker(\epsilon_R). \end{align*} This is the relative elementary subgroup under the stable convention used in the theorem. The plus construction requires a perfect [normal subgroup](/page/Normal%20Subgroup) of the fundamental group. Here $\pi_1(BGL(R))\cong GL(R)$, and [Stable Elementary Subgroup Is Normal][citetheorem:8655] gives $E(R)\trianglelefteq GL(R)$. The [Whitehead Lemma][citetheorem:8652] gives that $E(R)$ is perfect. Therefore $BGL(R)^+$ is the plus construction of $BGL(R)$ with respect to $E(R)$, and the same verification applies to $R/I$.[/guided]

[step:Identify kernel classes by stable matrices whose reductions are elementary]Let \begin{align*} q_R:BGL(R)\to BGL(R)^+ \end{align*} and \begin{align*} q_{R/I}:BGL(R/I)\to BGL(R/I)^+ \end{align*} be the plus-construction maps. By the chosen model, \begin{align*} K(R)=BGL(R)^+. \end{align*} By the defining fundamental-group property of the plus construction, applied to the perfect normal subgroup $E(R)\trianglelefteq GL(R)$ verified above, the plus construction with respect to $E(R)$ identifies \begin{align*} K_1(R)=\pi_1(BGL(R)^+)\cong GL(R)/E(R), \end{align*} and the same argument for $R/I$ gives \begin{align*} K_1(R/I)=\pi_1(BGL(R/I)^+)\cong GL(R/I)/E(R/I). \end{align*} Under these identifications, the homomorphism $K_1(R)\to K_1(R/I)$ is induced by \begin{align*} \alpha_R:GL(R)\to GL(R/I). \end{align*} This is because the map $BGL(R)\to BGL(R/I)$ induced by $\alpha_R$ carries $E(R)$ into $E(R/I)$, so it descends after plus construction to the quotient map on fundamental groups. Hence a class in \begin{align*} \ker\bigl(K_1(R)\to K_1(R/I)\bigr) \end{align*} is represented by an element $g\in GL(R)$ such that $\alpha_R(g)$ is trivial in $GL(R/I)/E(R/I)$. Equivalently, \begin{align*} \alpha_R(g)\in E(R/I). \end{align*} The long exact sequence for the homotopy fibre $K(R,I)$ gives \begin{align*} K_1(R,I)\to K_1(R)\to K_1(R/I), \end{align*} so this kernel is the image of $K_1(R,I)\to K_1(R)$. We do not identify it with all of $K_1(R,I)$; the possible preceding boundary from $K_2(R/I)$ is precisely why the corrected statement is formulated in terms of the kernel.[/step]

[guided]The point of this step is to use only the part of the homotopy-fibre exact sequence that is valid in degree one. Let $K(R,I)$ denote the homotopy fibre of $K(R)\to K(R/I)$. The long exact sequence of homotopy groups contains \begin{align*} K_2(R/I)\to K_1(R,I)\to K_1(R)\to K_1(R/I). \end{align*} Therefore $K_1(R,I)$ need not equal the kernel of $K_1(R)\to K_1(R/I)$; only its image in $K_1(R)$ is that kernel. This is the exact place where the possible $K_2(R/I)$ boundary enters. Now compute that kernel. The plus construction of $BGL(R)$ with respect to the verified perfect normal subgroup $E(R)\trianglelefteq GL(R)$ kills exactly $E(R)$ on fundamental groups, so \begin{align*} K_1(R)=\pi_1(BGL(R)^+)\cong GL(R)/E(R). \end{align*} Likewise, \begin{align*} K_1(R/I)=\pi_1(BGL(R/I)^+)\cong GL(R/I)/E(R/I). \end{align*} The quotient map $R\to R/I$ induces the stable group homomorphism \begin{align*} \alpha_R:GL(R)\to GL(R/I), \end{align*} and it sends elementary transvections over $R$ to elementary transvections over $R/I$. Hence it sends $E(R)$ into $E(R/I)$ and induces the displayed map on the quotient fundamental groups. Thus a class of $K_1(R)$ represented by $g\in GL(R)$ lies in the kernel exactly when the class of $\alpha_R(g)$ in $GL(R/I)/E(R/I)$ is the identity. This condition is \begin{align*} \alpha_R(g)\in E(R/I). \end{align*} This gives a matrix-level description of the kernel, not of the whole homotopy-fibre group.[/guided]

[step:Move each fibre-loop representative into the congruence subgroup] Let $g\in GL(R)$ satisfy $\alpha_R(g)\in E(R/I)$. We first justify that $\epsilon_R:E(R)\to E(R/I)$ is surjective. Every generator of $E(R/I)$ has the form $e_{ij}(\bar a)$ for some $\bar a\in R/I$; choose $a\in R$ mapping to $\bar a$, and then $e_{ij}(a)\in E(R)$ maps to $e_{ij}(\bar a)$. Since $E(R/I)$ is generated by such elementary transvections and $\epsilon_R$ is a group homomorphism, every finite product of generators in $E(R/I)$ lifts to a finite product in $E(R)$. Hence there exists an element $e\in E(R)$ such that \begin{align*} \epsilon_R(e)=\alpha_R(g). \end{align*} Define \begin{align*} h:=ge^{-1}\in GL(R). \end{align*} Then \begin{align*} \alpha_R(h)=\alpha_R(g)\alpha_R(e)^{-1}=1, \end{align*} so $h\in GL(R,I)$. The classes represented by $g$ and $h$ in $K_1(R)=GL(R)/E(R)$ are equal because $e\in E(R)$. Thus every element of $\ker\bigl(K_1(R)\to K_1(R/I)\bigr)$ has a representative in $GL(R,I)$. [/step]

[step:Identify exactly when two congruence representatives give the same kernel class] Let $h_0,h_1\in GL(R,I)$. Their images define the same element of $K_1(R)=GL(R)/E(R)$ precisely when \begin{align*} h_1h_0^{-1}\in E(R). \end{align*} Since $h_0,h_1\in GL(R,I)$ and $GL(R,I)$ is a subgroup, we also have \begin{align*} h_1h_0^{-1}\in GL(R,I). \end{align*} Therefore \begin{align*} h_1h_0^{-1}\in E(R)\cap GL(R,I)=\ker(E(R)\to E(R/I))=E(R,I). \end{align*} Conversely, if $h_1h_0^{-1}\in E(R,I)$, then $h_1h_0^{-1}\in E(R)$, so $h_0$ and $h_1$ have the same class in $GL(R)/E(R)=K_1(R)$. Because both representatives lie in $GL(R,I)$, both classes lie in the kernel of $K_1(R)\to K_1(R/I)$. Therefore the [equivalence relation](/page/Equivalence%20Relation) on congruence representatives for the kernel is precisely quotienting by $E(R,I)$. [/step]

[step:Construct the natural isomorphism]Define \begin{align*} \Phi:GL(R,I)\to \ker\bigl(K_1(R)\to K_1(R/I)\bigr) \end{align*} by sending $h\in GL(R,I)$ to its class $[h]$ in \begin{align*} K_1(R)=GL(R)/E(R). \end{align*} This is well-defined because $\alpha_R(h)=1$, so the image of $[h]$ in $GL(R/I)/E(R/I)=K_1(R/I)$ is the identity class. Multiplication of stable matrices descends to multiplication in $GL(R)/E(R)$, so $\Phi$ is a group homomorphism. The previous step shows that \begin{align*} \ker(\Phi)=E(R,I). \end{align*} The representative-moving step shows that $\Phi$ is surjective onto the kernel. Hence $\Phi$ induces a bijective group homomorphism \begin{align*} \overline{\Phi}:GL(R,I)/E(R,I)\to \ker\bigl(K_1(R)\to K_1(R/I)\bigr). \end{align*} Thus \begin{align*} GL(R,I)/E(R,I)\cong \ker\bigl(K_1(R)\to K_1(R/I)\bigr). \end{align*} By exactness of the low-degree sequence for the homotopy fibre, this kernel is the image of $K_1(R,I)\to K_1(R)$. The construction uses only the quotient homomorphism $R\to R/I$, stabilization, elementary subgroups, and the plus-construction identifications of $K_1$. Therefore the isomorphism is natural in pairs $(R,I)$.[/step]

[guided]We now package the matrix calculation into the promised quotient isomorphism. Define a map \begin{align*} \Phi:GL(R,I)\to \ker\bigl(K_1(R)\to K_1(R/I)\bigr) \end{align*} by sending $h\in GL(R,I)$ to its class $[h]$ in $K_1(R)=GL(R)/E(R)$. This lands in the kernel because $h\in GL(R,I)$ means $\alpha_R(h)=1$, and therefore the image of $[h]$ in $K_1(R/I)=GL(R/I)/E(R/I)$ is the identity class. The map $\Phi$ is a group homomorphism because multiplication in $GL(R,I)$ is the restriction of multiplication in $GL(R)$, and the quotient map $GL(R)\to GL(R)/E(R)$ is a group homomorphism. Its kernel consists exactly of those $h\in GL(R,I)$ whose class in $GL(R)/E(R)$ is the identity. This condition is $h\in E(R)$, and since $h\in GL(R,I)$ already, it is equivalent to \begin{align*} h\in E(R)\cap GL(R,I)=E(R,I). \end{align*} Thus $\ker(\Phi)=E(R,I)$. It remains to check surjectivity onto the kernel. Let a kernel class be represented by $g\in GL(R)$. By the kernel computation above, $\alpha_R(g)\in E(R/I)$. The restriction $\epsilon_R:E(R)\to E(R/I)$ is surjective: every elementary generator $e_{ij}(\bar a)$ over $R/I$ lifts to $e_{ij}(a)$ for any lift $a\in R$ of $\bar a$, and finite products of such generators lift to finite products in $E(R)$. Hence choose $e\in E(R)$ with $\epsilon_R(e)=\alpha_R(g)$ and set $h:=ge^{-1}$. Then \begin{align*} \alpha_R(h)=\alpha_R(g)\alpha_R(e)^{-1}=1, \end{align*} so $h\in GL(R,I)$. Since $e\in E(R)$, the classes of $g$ and $h$ in $GL(R)/E(R)$ are equal. Therefore every kernel class is in the image of $\Phi$. The [first isomorphism theorem](/theorems/791) now gives a bijective group homomorphism \begin{align*} \overline{\Phi}:GL(R,I)/E(R,I)\to \ker\bigl(K_1(R)\to K_1(R/I)\bigr). \end{align*} The low-degree exact sequence [Low Degree Relative Exact Sequence][citetheorem:8665] identifies this kernel with the image of $K_1(R,I)\to K_1(R)$, not necessarily with all of $K_1(R,I)$. Naturality follows because each ingredient in the construction is functorial in a morphism of pairs: the quotient map, the stabilized general linear group, elementary transvections, and the plus-construction identification of $K_1$.[/guided]