[proofplan]
We use the presentation $K_1(R)=GL(R)/E(R)$, where $GL(R)$ is the stable general linear group and $E(R)$ is the normal stable elementary subgroup. A unital ring homomorphism sends invertible matrices to invertible matrices by applying the homomorphism to an inverse, and this construction is compatible with stabilization. It also sends elementary transvections over $R$ to elementary transvections over $S$, so the stable map descends to the quotient. The identity and composition laws follow from entrywise equality before passing to the quotient.
[/proofplan]
[step:Apply $\varphi$ entrywise to define maps on finite general linear groups]
For each $n\in\mathbb N$, let $M_n(R)$ and $M_n(S)$ denote the rings of $n\times n$ matrices over $R$ and $S$, and let $GL_n(R)=M_n(R)^\times$ and $GL_n(S)=M_n(S)^\times$ denote their groups of units. Define
\begin{align*}
\varphi_n:M_n(R)\to M_n(S)
\end{align*}
by sending a matrix $A=(a_{ij})_{1\le i,j\le n}$ to the matrix
\begin{align*}
\varphi_n(A)=(\varphi(a_{ij}))_{1\le i,j\le n}.
\end{align*}
Since $\varphi$ is a unital ring homomorphism, $\varphi_n$ is a unital ring homomorphism and satisfies
\begin{align*}
\varphi_n(I_n)=I_n.
\end{align*}
If $A\in GL_n(R)$, let $B\in GL_n(R)$ denote its inverse, so $AB=BA=I_n$. Applying $\varphi_n$ gives
\begin{align*}
\varphi_n(A)\varphi_n(B)=\varphi_n(AB)=\varphi_n(I_n)=I_n
\end{align*}
and
\begin{align*}
\varphi_n(B)\varphi_n(A)=\varphi_n(BA)=\varphi_n(I_n)=I_n.
\end{align*}
Thus $\varphi_n(A)\in GL_n(S)$ with inverse $\varphi_n(B)$. Therefore the restriction
\begin{align*}
GL_n(\varphi):GL_n(R)\to GL_n(S)
\end{align*}
is a [group homomorphism](/page/Group%20Homomorphism).
[guided]
The first point is to check that entrywise application actually lands in invertible matrices. For each $n\in\mathbb N$, let $M_n(R)$ and $M_n(S)$ denote the rings of $n\times n$ matrices over $R$ and $S$, and let $GL_n(R)=M_n(R)^\times$ and $GL_n(S)=M_n(S)^\times$ denote their groups of units. Define the entrywise map
\begin{align*}
\varphi_n:M_n(R)\to M_n(S)
\end{align*}
by
\begin{align*}
\varphi_n((a_{ij})_{1\le i,j\le n})=(\varphi(a_{ij}))_{1\le i,j\le n}.
\end{align*}
Matrix addition and matrix multiplication are defined by finite sums and products of entries, and $\varphi$ preserves addition and multiplication. Since $\varphi$ is unital, it also sends $1_R$ to $1_S$, so $\varphi_n$ sends the identity matrix $I_n\in M_n(R)$ to the identity matrix $I_n\in M_n(S)$.
Now let $A\in GL_n(R)$. By definition, there is a matrix $B\in GL_n(R)$ such that $AB=BA=I_n$. Applying the ring homomorphism $\varphi_n$ to both products gives
\begin{align*}
\varphi_n(A)\varphi_n(B)=\varphi_n(AB)=\varphi_n(I_n)=I_n
\end{align*}
and
\begin{align*}
\varphi_n(B)\varphi_n(A)=\varphi_n(BA)=\varphi_n(I_n)=I_n.
\end{align*}
Thus $\varphi_n(B)$ is a two-sided inverse for $\varphi_n(A)$ in $M_n(S)$, so $\varphi_n(A)\in GL_n(S)$. Therefore the restriction
\begin{align*}
GL_n(\varphi):GL_n(R)\to GL_n(S)
\end{align*}
is well-defined. It is a group homomorphism because multiplication in $GL_n(R)$ and $GL_n(S)$ is matrix multiplication and $\varphi_n$ preserves matrix multiplication.
[/guided]
[/step]
[step:Pass the finite-level maps to the stable general linear group]
Let
\begin{align*}
\sigma_{n,R}:GL_n(R)\to GL_{n+1}(R)
\end{align*}
denote the stabilization homomorphism
\begin{align*}
A\mapsto \operatorname{diag}(A,1_R),
\end{align*}
and define $\sigma_{n,S}:GL_n(S)\to GL_{n+1}(S)$ analogously. For every $A=(a_{ij})_{1\le i,j\le n}\in GL_n(R)$, unitality of $\varphi$ gives
\begin{align*}
GL_{n+1}(\varphi)(\operatorname{diag}(A,1_R))=\operatorname{diag}(GL_n(\varphi)(A),1_S).
\end{align*}
Equivalently,
\begin{align*}
GL_{n+1}(\varphi)\circ\sigma_{n,R}=\sigma_{n,S}\circ GL_n(\varphi).
\end{align*}
Hence the maps $GL_n(\varphi)$ are compatible with the directed systems defining the stable groups. Here
\begin{align*}
GL(R)=\varinjlim_n GL_n(R)
\end{align*}
and
\begin{align*}
GL(S)=\varinjlim_n GL_n(S)
\end{align*}
are the direct limits taken with respect to the stabilization homomorphisms. By the universal property of the direct limit of groups, the compatible finite-level maps induce a unique group homomorphism
\begin{align*}
GL(\varphi):GL(R)\to GL(S)
\end{align*}
given by entrywise application of $\varphi$ to any representative finite matrix.
[/step]
[step:Check that elementary matrices map to elementary matrices]
For $n\ge 2$, distinct indices $1\le i\ne j\le n$, and an element $r\in R$, let
\begin{align*}
e_{ij}(r)=I_n+rE_{ij}\in GL_n(R)
\end{align*}
denote the elementary transvection, where $0_R$ is the zero element of $R$ and $E_{ij}\in M_n(R)$ is the matrix with $1_R$ in the $(i,j)$-entry and $0_R$ elsewhere. Applying $\varphi_n$ entrywise gives
\begin{align*}
GL_n(\varphi)(e_{ij}(r))=I_n+\varphi(r)E_{ij}=e_{ij}(\varphi(r))\in GL_n(S).
\end{align*}
The stable elementary subgroup $E(R)\le GL(R)$ is generated by the images of all such elementary transvections. Since $GL(\varphi)$ sends each generator of $E(R)$ into $E(S)$, it sends the subgroup generated by them into $E(S)$:
\begin{align*}
GL(\varphi)(E(R))\subset E(S).
\end{align*}
[/step]
[step:Descend the stable map to the quotient defining $K_1$]
Using the definition
\begin{align*}
K_1(R)=GL(R)/E(R)
\end{align*}
and similarly
\begin{align*}
K_1(S)=GL(S)/E(S),
\end{align*}
where $E(R)\trianglelefteq GL(R)$ and $E(S)\trianglelefteq GL(S)$ are the normal stable elementary subgroups appearing in the definition of $K_1$, define
\begin{align*}
K_1(\varphi):K_1(R)\to K_1(S)
\end{align*}
by
\begin{align*}
[A]\mapsto [GL(\varphi)(A)],
\end{align*}
where $A\in GL(R)$ and brackets denote the corresponding coset modulo the stable elementary subgroup.
This definition is independent of the representative. Indeed, suppose $A,A'\in GL(R)$ represent the same element of $K_1(R)$. Then there exists $e\in E(R)$ such that
\begin{align*}
A'=Ae.
\end{align*}
Applying $GL(\varphi)$ gives
\begin{align*}
GL(\varphi)(A')=GL(\varphi)(A)GL(\varphi)(e).
\end{align*}
By the previous step, $GL(\varphi)(e)\in E(S)$, so $GL(\varphi)(A')$ and $GL(\varphi)(A)$ determine the same coset in $GL(S)/E(S)$. Therefore $K_1(\varphi)$ is well-defined. Since $GL(\varphi)$ is a group homomorphism, the induced map on quotients is also a group homomorphism.
[/step]
[step:Verify identity and composition before passing to quotients]
Let $R$ be a unital ring. For every $n\in\mathbb N$ and every $A=(a_{ij})_{1\le i,j\le n}\in GL_n(R)$, entrywise application of $\operatorname{id}_R$ gives
\begin{align*}
GL_n(\operatorname{id}_R)(A)=A.
\end{align*}
Passing to the stable group and then to the quotient gives
\begin{align*}
K_1(\operatorname{id}_R)=\operatorname{id}_{K_1(R)}.
\end{align*}
Now let $\varphi:R\to S$ and $\psi:S\to T$ be unital ring homomorphisms. For every $n\in\mathbb N$ and every $A=(a_{ij})_{1\le i,j\le n}\in GL_n(R)$,
\begin{align*}
GL_n(\psi\circ\varphi)(A)=((\psi\circ\varphi)(a_{ij}))_{1\le i,j\le n}.
\end{align*}
The right-hand side is exactly
\begin{align*}
GL_n(\psi)(GL_n(\varphi)(A)).
\end{align*}
Thus
\begin{align*}
GL(\psi\circ\varphi)=GL(\psi)\circ GL(\varphi).
\end{align*}
Since both sides send elementary subgroups into elementary subgroups, passing to quotients gives
\begin{align*}
K_1(\psi\circ\varphi)=K_1(\psi)\circ K_1(\varphi).
\end{align*}
This proves the functoriality assertions.
[/step]
