[proofplan] For $K_0$, we extend scalars: a finitely generated projective left $R$-module $P$ is sent to the finitely generated projective left $S$-module $S\otimes_R P$, where $S$ is regarded as an $(S,R)$-bimodule through $\varphi$. This gives a homomorphism on the commutative monoid $V(R)$ of finite projective classes, and the universal property of group completion extends it uniquely to $K_0(R)$. For $K_1$, we apply $\varphi$ entrywise to invertible matrices, pass to the stable general linear group, and use the fact that elementary matrices are sent to elementary matrices. The identity and composition laws then follow from the canonical [tensor product](/page/Tensor%20Product) isomorphisms for $K_0$ and from entrywise functoriality for $K_1$. [/proofplan] [step:Extend scalars from finite projective $R$-modules to finite projective $S$-modules] Regard $S$ as an $(S,R)$-bimodule by left multiplication in $S$ and right action \begin{align*} s\cdot r=s\varphi(r) \end{align*} for $s\in S$ and $r\in R$. For every left $R$-module $P$, define the left $S$-module \begin{align*} \varphi_*P:=S\otimes_R P, \end{align*} where $S$ acts on the left tensor factor. Suppose $P$ is a finitely generated projective left $R$-module. By [citetheorem:8628], there exist an integer $n\in\mathbb N$ and a left $R$-module $Q$ such that \begin{align*} P\oplus Q\cong R^n. \end{align*} Tensoring this isomorphism over $R$ with $S$ gives an isomorphism of left $S$-modules \begin{align*} (S\otimes_R P)\oplus(S\otimes_R Q)\cong S\otimes_R R^n. \end{align*} The map \begin{align*} S\otimes_R R^n&\to S^n \end{align*} defined on elementary tensors by \begin{align*} s\otimes(r_1,\dots,r_n)&\mapsto (s\varphi(r_1),\dots,s\varphi(r_n)) \end{align*} is an isomorphism of left $S$-modules. Hence \begin{align*} (S\otimes_R P)\oplus(S\otimes_R Q)\cong S^n. \end{align*} Again by [citetheorem:8628], $S\otimes_R P$ is finitely generated projective as a left $S$-module. [guided] The first point is to make the tensor product meaningful for possibly noncommutative rings. Since $P$ is a left $R$-module and $\varphi:R\to S$ is a unital ring homomorphism, we make $S$ into a right $R$-module by declaring \begin{align*} s\cdot r=s\varphi(r) \end{align*} for $s\in S$ and $r\in R$. Together with the ordinary left multiplication of $S$ on itself, this makes $S$ an $(S,R)$-bimodule. Therefore $S\otimes_R P$ is defined, and its left $S$-module structure is given by multiplication on the first tensor factor. We now verify that finite projectivity is preserved. Let $P$ be a finitely generated projective left $R$-module. By [citetheorem:8628], there are an integer $n\in\mathbb N$ and a left $R$-module $Q$ with \begin{align*} P\oplus Q\cong R^n. \end{align*} Tensoring over $R$ with $S$ preserves direct sums, so the preceding isomorphism gives an isomorphism of left $S$-modules \begin{align*} (S\otimes_R P)\oplus(S\otimes_R Q)\cong S\otimes_R R^n. \end{align*} The module $S\otimes_R R^n$ is naturally the free left $S$-module $S^n$: the explicit isomorphism is the map \begin{align*} S\otimes_R R^n&\to S^n \end{align*} given on elementary tensors by \begin{align*} s\otimes(r_1,\dots,r_n)&\mapsto (s\varphi(r_1),\dots,s\varphi(r_n)). \end{align*} Its inverse sends $(s_1,\dots,s_n)\in S^n$ to \begin{align*} \sum_{i=1}^n s_i\otimes e_i, \end{align*} where $e_i\in R^n$ denotes the $i$-th standard basis vector. Thus \begin{align*} (S\otimes_R P)\oplus(S\otimes_R Q)\cong S^n. \end{align*} Applying [citetheorem:8628] over the ring $S$, this shows that $S\otimes_R P$ is finitely generated projective as a left $S$-module. This is exactly the condition needed for extension of scalars to define a map on $K_0$-classes. [/guided] [/step] [step:Pass from the monoid of projective classes to $K_0$] Let $V(R)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $R$-modules under direct sum. Define \begin{align*} V(\varphi):V(R)&\to V(S) \end{align*} by \begin{align*} [P]&\mapsto [S\otimes_R P]. \end{align*} This is well-defined because an $R$-linear isomorphism $P\cong P'$ induces an $S$-linear isomorphism $S\otimes_R P\cong S\otimes_R P'$. It is a monoid homomorphism because the canonical map \begin{align*} (S\otimes_R P)\oplus(S\otimes_R Q)&\to S\otimes_R(P\oplus Q) \end{align*} is an isomorphism of left $S$-modules, and $S\otimes_R 0\cong 0$. By the universal property of group completion [citetheorem:8633], the monoid homomorphism $V(\varphi):V(R)\to V(S)$ induces a unique [group homomorphism](/page/Group%20Homomorphism) \begin{align*} K_0(\varphi):K_0(R)\to K_0(S) \end{align*} satisfying \begin{align*} K_0(\varphi)([P])=[S\otimes_R P] \end{align*} for every finitely generated projective left $R$-module $P$. [/step] [step:Verify identity and composition for $K_0$ using tensor product isomorphisms] For the identity homomorphism $\operatorname{id}_R:R\to R$, the map \begin{align*} R\otimes_R P&\to P \end{align*} given by \begin{align*} r\otimes p&\mapsto rp \end{align*} is an isomorphism of left $R$-modules for every left $R$-module $P$. Hence \begin{align*} K_0(\operatorname{id}_R)([P])=[P], \end{align*} so \begin{align*} K_0(\operatorname{id}_R)=\operatorname{id}_{K_0(R)}. \end{align*} Now let \begin{align*} R\xrightarrow{\varphi}S\xrightarrow{\psi}T \end{align*} be homomorphisms of unital rings. For every left $R$-module $P$, the map \begin{align*} T\otimes_S(S\otimes_R P)&\to T\otimes_R P \end{align*} defined by \begin{align*} t\otimes(s\otimes p)&\mapsto t\psi(s)\otimes p \end{align*} is an isomorphism of left $T$-modules, where $T$ is regarded as a right $R$-module through $\psi\circ\varphi$. Therefore, for every finitely generated projective left $R$-module $P$, \begin{align*} K_0(\psi)(K_0(\varphi)([P]))=[T\otimes_S(S\otimes_R P)]=[T\otimes_R P]=K_0(\psi\circ\varphi)([P]). \end{align*} Since $K_0(R)$ is generated as an abelian group by classes $[P]$ of finitely generated projective left $R$-modules, it follows that \begin{align*} K_0(\psi\circ\varphi)=K_0(\psi)\circ K_0(\varphi). \end{align*} [/step] [step:Apply $\varphi$ entrywise to stable invertible matrices] For each $n\in\mathbb N$, let $GL_n(R)$ denote the group of invertible $n\times n$ matrices over $R$. The stabilization map $GL_n(R)\to GL_{n+1}(R)$ sends a matrix $A$ to the block diagonal matrix with diagonal blocks $A$ and $1_R$. Define the stable general linear group by \begin{align*} GL(R):=\varinjlim_n GL_n(R). \end{align*} For $i\ne j$ and $r\in R$, let $e_{ij}(r)\in GL_n(R)$ denote the elementary matrix with diagonal entries $1_R$, off-diagonal $(i,j)$-entry $r$, and all other off-diagonal entries $0_R$. Let $E(R)\le GL(R)$ be the subgroup generated by the images of all such elementary matrices under the structure maps $GL_n(R)\to GL(R)$. Now define \begin{align*} GL_n(\varphi):GL_n(R)&\to GL_n(S) \end{align*} by applying $\varphi$ to each matrix entry. Since $\varphi$ is a unital ring homomorphism, matrix multiplication and identity matrices are preserved, so $GL_n(\varphi)$ is a group homomorphism. Entrywise application of $\varphi$ commutes with stabilization because $\varphi(1_R)=1_S$, so the maps $GL_n(\varphi)$ induce a homomorphism on the colimit \begin{align*} GL(\varphi):GL(R)\to GL(S). \end{align*} Finally, $GL_n(\varphi)(e_{ij}(r))=e_{ij}(\varphi(r))$, so $GL(\varphi)$ carries $E(R)$ into $E(S)$. [/step] [step:Descend the stable matrix map to $K_1$] Using the definition \begin{align*} K_1(R)=GL(R)/E(R), \end{align*} the inclusion \begin{align*} GL(\varphi)(E(R))\subseteq E(S) \end{align*} implies that $GL(\varphi)$ descends to a group homomorphism \begin{align*} K_1(\varphi):K_1(R)&\to K_1(S) \end{align*} given by \begin{align*} [A]_{K_1(R)}&\mapsto [\varphi(A)]_{K_1(S)}. \end{align*} Here $A\in GL_n(R)$ for some $n\in\mathbb N$, $\varphi(A)\in GL_n(S)$ is obtained by applying $\varphi$ entrywise, and the brackets denote the corresponding cosets modulo the stable elementary subgroups. For the identity homomorphism $\operatorname{id}_R$, entrywise application is the identity on every $GL_n(R)$, hence \begin{align*} K_1(\operatorname{id}_R)=\operatorname{id}_{K_1(R)}. \end{align*} For composable homomorphisms \begin{align*} R\xrightarrow{\varphi}S\xrightarrow{\psi}T, \end{align*} entrywise application satisfies \begin{align*} (\psi\circ\varphi)(A)=\psi(\varphi(A)) \end{align*} for every stable invertible matrix $A\in GL(R)$. Passing to cosets modulo elementary subgroups gives \begin{align*} K_1(\psi\circ\varphi)=K_1(\psi)\circ K_1(\varphi). \end{align*} Together with the $K_0$ construction above, this proves functoriality of both low algebraic $K$-groups. [/step]