[proofplan] We first construct the algebraic direct limit by declaring two elements from possibly different stages equivalent when they agree after passing to a common later stage. The algebraic operations are defined by moving representatives to a common upper bound, and the norm is obtained as the eventual infimum of the norms of later images. This seminorm satisfies the $C^*$-identity, so quotienting by its zero space and completing gives a $C^*$-algebra. The universal property follows by factoring any compatible family through the algebraic limit and using contractivity to extend continuously. Finally, when the connecting maps are injective, they are isometric by uniqueness of the $C^*$-norm, and the limit is the closure of the union of the embedded stages. [/proofplan]

[step:Form the algebraic direct limit by eventual equality] Let \begin{align*} S:=\bigsqcup_{i\in I} A_i \end{align*} be the disjoint union of the underlying sets of the algebras. For $a\in A_i$ and $b\in A_j$, define $(i,a)\sim (j,b)$ if there exists $k\in I$ with $i\leq k$ and $j\leq k$ such that \begin{align*} \varphi_{ik}(a)=\varphi_{jk}(b) \end{align*} in $A_k$. This relation is an [equivalence relation](/page/Equivalence%20Relation). Reflexivity follows from $\varphi_{ii}=\operatorname{id}_{A_i}$. Symmetry is immediate from equality. For transitivity, suppose $(i,a)\sim (j,b)$ and $(j,b)\sim (\ell,c)$. Choose $k_1,k_2\in I$ such that $i,j\leq k_1$, $j,\ell\leq k_2$, \begin{align*} \varphi_{ik_1}(a)=\varphi_{jk_1}(b) \end{align*} and \begin{align*} \varphi_{jk_2}(b)=\varphi_{\ell k_2}(c). \end{align*} Since $I$ is directed, choose $m\in I$ with $k_1\leq m$ and $k_2\leq m$. Applying the connecting maps into $A_m$ and using $\varphi_{k_1m}\circ\varphi_{ik_1}=\varphi_{im}$, $\varphi_{k_1m}\circ\varphi_{jk_1}=\varphi_{jm}$, $\varphi_{k_2m}\circ\varphi_{jk_2}=\varphi_{jm}$, and $\varphi_{k_2m}\circ\varphi_{\ell k_2}=\varphi_{\ell m}$ gives \begin{align*} \varphi_{im}(a)=\varphi_{\ell m}(c). \end{align*} Thus $(i,a)\sim(\ell,c)$. Define the algebraic direct-limit set \begin{align*} A_{\mathrm{alg}}:=S/\sim. \end{align*} For $a\in A_i$, denote its equivalence class by $[i,a]$. [/step]

[step:Define the $*$-algebra operations at a common upper stage] For $[i,a],[j,b]\in A_{\mathrm{alg}}$, choose $k\in I$ with $i\leq k$ and $j\leq k$, and define \begin{align*} [i,a]+[j,b]:=[k,\varphi_{ik}(a)+\varphi_{jk}(b)]. \end{align*} For $\lambda\in\mathbb C$, define \begin{align*} \lambda[i,a]:=[i,\lambda a]. \end{align*} Define multiplication and involution by \begin{align*} [i,a][j,b]:=[k,\varphi_{ik}(a)\varphi_{jk}(b)] \end{align*} for any common upper bound $k$ of $i$ and $j$, and \begin{align*} [i,a]^*:=[i,a^*]. \end{align*} These definitions are independent of the chosen common upper bound. Indeed, if $k$ and $k'$ are two common upper bounds of $i$ and $j$, choose $m\in I$ with $k\leq m$ and $k'\leq m$. Since the connecting maps are $*$-homomorphisms and satisfy the directed-system identities, applying $\varphi_{km}$ and $\varphi_{k'm}$ to the two proposed representatives gives the same element of $A_m$. They are also independent of representatives. For example, suppose $[i,a]=[i',a']$ and $[j,b]=[j',b']$. Choose $r\in I$ with $i,i'\leq r$ and \begin{align*} \varphi_{ir}(a)=\varphi_{i'r}(a'), \end{align*} and choose $s\in I$ with $j,j'\leq s$ and \begin{align*} \varphi_{js}(b)=\varphi_{j's}(b'). \end{align*} Choose $m\in I$ dominating $r$ and $s$. Then the images of $a,a',b,b'$ in $A_m$ satisfy \begin{align*} \varphi_{im}(a)=\varphi_{i'm}(a') \end{align*} and \begin{align*} \varphi_{jm}(b)=\varphi_{j'm}(b'). \end{align*} Therefore the sums, products, scalar multiples, and adjoints computed from either representatives have the same image in $A_m$, so they define the same class. With these operations, $A_{\mathrm{alg}}$ is a complex involutive algebra. The algebraic identities follow after moving all representatives appearing in the identity to a common upper stage and using the corresponding identity in that $C^*$-algebra. [/step]

[step:Define the canonical seminorm by decreasing later-stage norms]Define the function \begin{align*} p: A_{\mathrm{alg}} &\to [0,\infty), \qquad [i,a] \mapsto \inf\{\|\varphi_{ij}(a)\|_{A_j}:j\in I,\ i\leq j\}. \end{align*} Equivalently, $p([i,a])$ is the limit of the decreasing net $j\mapsto \|\varphi_{ij}(a)\|_{A_j}$ over the directed set $\{j\in I:i\leq j\}$. The net is decreasing because every connecting map is contractive by [citetheorem:8547]: if $i\leq j\leq k$, then \begin{align*} \|\varphi_{ik}(a)\|_{A_k}=\|\varphi_{jk}(\varphi_{ij}(a))\|_{A_k}\leq \|\varphi_{ij}(a)\|_{A_j}. \end{align*} The value of $p$ is independent of representatives. Suppose $[i,a]=[j,b]$. Choose $k\in I$ with $i,j\leq k$ and \begin{align*} \varphi_{ik}(a)=\varphi_{jk}(b). \end{align*} For every $m\in I$ with $k\leq m$, the directed-system identities give \begin{align*} \varphi_{im}(a)=\varphi_{jm}(b). \end{align*} Thus the two decreasing nets of norms agree on a common cofinal tail, and hence have the same infimum. Therefore $p$ is well-defined.[/step]

[guided]The point of the seminorm is to measure the size of an algebraic class only after allowing it to move arbitrarily far forward in the system. For $a\in A_i$, define \begin{align*} p([i,a]):=\inf\{\|\varphi_{ij}(a)\|_{A_j}:j\in I,\ i\leq j\}. \end{align*} This is the correct formula because later connecting maps may decrease norms when they have nontrivial kernels, so the norm of the class should be the eventual size that survives in the limit. We first verify that the numbers inside the infimum form a decreasing net. If $i\leq j\leq k$, then the compatibility of the directed system gives \begin{align*} \varphi_{ik}(a)=\varphi_{jk}(\varphi_{ij}(a)). \end{align*} The map $\varphi_{jk}:A_j\to A_k$ is a $*$-homomorphism between $C^*$-algebras. By [citetheorem:8547], it is contractive, so \begin{align*} \|\varphi_{ik}(a)\|_{A_k}\leq \|\varphi_{ij}(a)\|_{A_j}. \end{align*} Thus passing farther forward never increases the norm. Now we check that the formula does not depend on the representative. Suppose $[i,a]=[j,b]$. By definition of the equivalence relation, there is $k\in I$ with $i,j\leq k$ such that \begin{align*} \varphi_{ik}(a)=\varphi_{jk}(b). \end{align*} For every $m\in I$ with $k\leq m$, apply $\varphi_{km}:A_k\to A_m$ to this equality. The compatibility identities give \begin{align*} \varphi_{im}(a)=\varphi_{km}(\varphi_{ik}(a))=\varphi_{km}(\varphi_{jk}(b))=\varphi_{jm}(b). \end{align*} Therefore the two tails of later-stage norms agree exactly: \begin{align*} \|\varphi_{im}(a)\|_{A_m}=\|\varphi_{jm}(b)\|_{A_m} \end{align*} for all $m\geq k$. Since an infimum over a directed decreasing net is unchanged after passing to a cofinal tail, both representatives give the same value of $p$. Hence $p$ is a well-defined function on $A_{\mathrm{alg}}$.[/guided]

[step:Verify that the seminorm satisfies the $C^*$-identity] The function $p:A_{\mathrm{alg}}\to[0,\infty)$ is a seminorm. For $[i,a],[j,b]\in A_{\mathrm{alg}}$, choose $k\in I$ with $i,j\leq k$. Given $\varepsilon>0$, choose $m_1\geq k$ and $m_2\geq k$ so that \begin{align*} \|\varphi_{im_1}(a)\|_{A_{m_1}}\leq p([i,a])+\varepsilon \end{align*} and \begin{align*} \|\varphi_{jm_2}(b)\|_{A_{m_2}}\leq p([j,b])+\varepsilon. \end{align*} Choose a common upper bound $m\in I$ of $m_1$ and $m_2$. Since the two norm nets are decreasing, monotonicity gives \begin{align*} \|\varphi_{im}(a)\|_{A_m}\leq p([i,a])+\varepsilon \end{align*} and \begin{align*} \|\varphi_{jm}(b)\|_{A_m}\leq p([j,b])+\varepsilon. \end{align*} Applying the triangle inequality in $A_m$ yields \begin{align*} \|\varphi_{im}(a)+\varphi_{jm}(b)\|_{A_m}\leq p([i,a])+p([j,b])+2\varepsilon. \end{align*} Because $m\geq k$, this estimate applies to a representative of $[i,a]+[j,b]$, so \begin{align*} p([i,a]+[j,b])\leq p([i,a])+p([j,b])+2\varepsilon. \end{align*} Letting $\varepsilon\downarrow 0$ gives \begin{align*} p([i,a]+[j,b])\leq p([i,a])+p([j,b]). \end{align*} Homogeneity follows from homogeneity of the norms in each $A_j$. The same common-upper-bound argument proves submultiplicativity. For $m\geq k$, \begin{align*} \|\varphi_{im}(a)\varphi_{jm}(b)\|_{A_m}\leq \|\varphi_{im}(a)\|_{A_m}\|\varphi_{jm}(b)\|_{A_m}. \end{align*} Taking infima along the decreasing nets gives \begin{align*} p([i,a][j,b])\leq p([i,a])p([j,b]). \end{align*} Indeed, if $\alpha:=p([i,a])$ and $\beta:=p([j,b])$, then for every $\varepsilon>0$ there are $m_1,m_2\geq k$ with $\|\varphi_{im_1}(a)\|_{A_{m_1}}\leq \alpha+\varepsilon$ and $\|\varphi_{jm_2}(b)\|_{A_{m_2}}\leq \beta+\varepsilon$; choosing a common upper bound $m\geq m_1,m_2$ and using monotonicity gives a product bounded by $(\alpha+\varepsilon)(\beta+\varepsilon)$. Conversely, the infimum of nonnegative products is bounded below by $\alpha\beta$ because each factor is bounded below by its own infimum. Thus the common-upper-bound argument justifies taking the product of the two infima, and letting $\varepsilon\downarrow 0$ gives the displayed inequality. Also, \begin{align*} p([i,a]^*)=p([i,a]) \end{align*} because every connecting map preserves the involution and each $C^*$-norm satisfies $\|x^*\|=\|x\|$. Finally, for $[i,a]\in A_{\mathrm{alg}}$, \begin{align*} p([i,a]^*[i,a])=\inf_{j\geq i}\|\varphi_{ij}(a)^*\varphi_{ij}(a)\|_{A_j}. \end{align*} Using the $C^*$-identity in each $A_j$ gives \begin{align*} p([i,a]^*[i,a])=\inf_{j\geq i}\|\varphi_{ij}(a)\|_{A_j}^{2}. \end{align*} Since the net $j\mapsto \|\varphi_{ij}(a)\|_{A_j}$ is decreasing and nonnegative, \begin{align*} \inf_{j\geq i}\|\varphi_{ij}(a)\|_{A_j}^{2}=p([i,a])^2. \end{align*} Thus \begin{align*} p(x^*x)=p(x)^2 \end{align*} for every $x\in A_{\mathrm{alg}}$. [/step]

[step:Quotient by the zero seminorm and complete] Define \begin{align*} N:=\{x\in A_{\mathrm{alg}}:p(x)=0\}. \end{align*} Since $p$ is submultiplicative and $p(x^*)=p(x)$, the set $N$ is a self-adjoint two-sided ideal in $A_{\mathrm{alg}}$. Let \begin{align*} A_0:=A_{\mathrm{alg}}/N \end{align*} and define \begin{align*} \|x+N\|_{A_0}:=p(x). \end{align*} This is a well-defined norm on $A_0$, and the quotient involution and multiplication make $A_0$ a normed involutive algebra. The $C^*$-identity on $A_{\mathrm{alg}}$ descends to \begin{align*} \|(x+N)^*(x+N)\|_{A_0}=\|x+N\|_{A_0}^{2}. \end{align*} Let $A$ be the completion of the normed space $A_0$. The multiplication and involution on $A_0$ are norm-continuous because multiplication is submultiplicative and the involution is isometric, so they extend uniquely to continuous multiplication and involution on $A$. The algebraic identities for multiplication, distributivity, scalar multiplication, and involution remain valid on $A$: both sides of each identity are continuous functions of the variables and agree on the dense subalgebra $A_0$. In particular, the extended multiplication is associative and distributive, and the extended involution is conjugate-linear, antimultiplicative, and satisfies $(x^*)^*=x$. The $C^*$-identity extends from $A_0$ to $A$ by taking limits of Cauchy sequences. Hence $A$ is a $C^*$-algebra. For each $i\in I$, define the canonical map \begin{align*} \varphi_i:A_i\to A \end{align*} as the composition \begin{align*} a\mapsto [i,a]\mapsto [i,a]+N\mapsto [i,a]+N\in A. \end{align*} Because the algebraic operations on $A_{\mathrm{alg}}$ were defined stagewise, each $\varphi_i$ is a $*$-homomorphism. If $i\leq j$, then $[i,a]=[j,\varphi_{ij}(a)]$ for every $a\in A_i$, and therefore \begin{align*} \varphi_j(\varphi_{ij}(a))=\varphi_i(a). \end{align*} Thus the canonical maps are compatible with the directed system. [/step]

[step:Prove the universal property by continuous extension] Let $B$ be a $C^*$-algebra, and let $(\psi_i)_{i\in I}$ be a compatible family of $*$-homomorphisms $\psi_i:A_i\to B$, so that $\psi_j\circ\varphi_{ij}=\psi_i$ whenever $i\leq j$. Define an algebraic map \begin{align*} \psi_{\mathrm{alg}}:A_{\mathrm{alg}}\to B \end{align*} by \begin{align*} \psi_{\mathrm{alg}}([i,a]):=\psi_i(a). \end{align*} This is well-defined: if $[i,a]=[j,b]$, choose $k\in I$ with $i,j\leq k$ and $\varphi_{ik}(a)=\varphi_{jk}(b)$. Then compatibility gives \begin{align*} \psi_i(a)=\psi_k(\varphi_{ik}(a))=\psi_k(\varphi_{jk}(b))=\psi_j(b). \end{align*} The map $\psi_{\mathrm{alg}}$ is a $*$-homomorphism because addition, multiplication, scalar multiplication, and involution in $A_{\mathrm{alg}}$ are computed at a common upper stage, where the corresponding $\psi_k$ is a $*$-homomorphism. For $a\in A_i$ and every $j\geq i$, compatibility gives \begin{align*} \psi_i(a)=\psi_j(\varphi_{ij}(a)). \end{align*} By contractivity of $*$-homomorphisms from [citetheorem:8547], \begin{align*} \|\psi_i(a)\|_B\leq \|\varphi_{ij}(a)\|_{A_j}. \end{align*} Taking the infimum over all $j\geq i$ yields \begin{align*} \|\psi_{\mathrm{alg}}([i,a])\|_B\leq p([i,a]). \end{align*} Therefore $\psi_{\mathrm{alg}}$ vanishes on $N$ and descends to a contractive $*$-homomorphism \begin{align*} \psi_0:A_0\to B. \end{align*} Since $A$ is the completion of $A_0$ and $\psi_0$ is contractive, there is a unique continuous [linear map](/page/Linear%20Map) \begin{align*} \psi:A\to B \end{align*} extending $\psi_0$. The multiplicative and involutive identities pass to the extension by continuity and density of $A_0$ in $A$, so $\psi$ is a $*$-homomorphism. By construction, \begin{align*} \psi\circ\varphi_i=\psi_i \end{align*} for every $i\in I$. Uniqueness follows because the algebraic span of $\bigcup_{i\in I}\varphi_i(A_i)$ is dense in $A$. Any continuous $*$-homomorphism $A\to B$ agreeing with all $\psi_i$ on the images of the stages agrees with $\psi_0$ on $A_0$, and hence agrees with $\psi$ on the completion. [/step]

[step:Identify the injective case with the closure of the increasing union] Assume now that every connecting map $\varphi_{ij}$ is injective. We use the standard faithfulness-isometry theorem for $C^*$-algebras: an injective $*$-homomorphism between $C^*$-algebras is isometric. The hypotheses match the present situation because, for each $i\leq j$, both $A_i$ and $A_j$ are $C^*$-algebras and $\varphi_{ij}:A_i\to A_j$ is an injective $*$-homomorphism by assumption. Therefore \begin{align*} \|\varphi_{ij}(a)\|_{A_j}=\|a\|_{A_i} \end{align*} for every $a\in A_i$. Thus every connecting map is isometric. It follows that for $a\in A_i$, \begin{align*} p([i,a])=\inf_{j\geq i}\|\varphi_{ij}(a)\|_{A_j}=\|a\|_{A_i}. \end{align*} Therefore $\varphi_i(a)=0$ implies $p([i,a])=0$, hence $\|a\|_{A_i}=0$, and so $a=0$. Thus each canonical map $\varphi_i:A_i\to A$ is injective and isometric. Finally, by construction, $A_0$ is the algebraic union of the images of the stages after the identifications supplied by the maps $\varphi_i$. More precisely, every element of $A_{\mathrm{alg}}$ has the form $[i,a]$, and its image in $A_0$ is $\varphi_i(a)$. Hence \begin{align*} A_0=\bigcup_{i\in I}\varphi_i(A_i) \end{align*} inside $A$, with the union understood through the compatible inclusions. Since $A$ is the norm completion of $A_0$, we obtain \begin{align*} A=\overline{\bigcup_{i\in I}\varphi_i(A_i)}^{\|\cdot\|_A}. \end{align*} This proves the injective statement and completes the construction of the $C^*$-inductive limit. [/step]