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I want to realize the inductive limit of a system of $C^*$-algebras, say $\{\phi_{ij} : A_i \to A_j \}_{i \leq j}$ as a sub $C^*$-algebra of
$$ Q := \frac{ \prod_i A_i }{ \sum_i A_i}. $$ The index set should be an arbitrary directed set $(I,\leq)$. By $\prod_i A_i$ I mean bounded sequences $(a_i)_{i \in I}$ with $a_i \in A_i$. By $\sum_i A_i$ I mean the sequences $(a_i)_i$ such that $\lVert a_i \rVert \to 0$ along the filter $I_{\mathrm{cofin}}$.

Analogously to the case $I = \mathbb N$ one may define maps $\psi_i : A_i \to Q$ by $$ \psi_i(x) = \pi((a_j)_j), $$ where $a_j = \phi_{ij}(x)$ if $j \geq i$ and $0$ otherwise. However, I am not sure if $$ A := \overline{\bigcup_i \psi_i(A_i)} $$ is the inductive limit of the $A_i$, since I cannot prove that $\psi_j \circ \phi_{ij} = \psi_i$. Maybe this construction does not work at all and there is another way to realize the inductive limit as a sub $C^*$-algebra of $Q$.

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The algebra $A$ you have constructed is the inductive limit. To show that $\psi_j\circ \phi_{i,j} = \psi_i$, choose $a \in A_j$, then $$ \psi_i(a) - \psi_j\circ\phi_{i,j}(a)) = \pi(b_j) $$ where $b_k = 0$ if $k\geq j$. This is clearly an element of $\sum A_j$, so $\pi(b_j) = 0$.


To show that this algebra is in the inductive limit, note that if $a\in A_i$, then $$ \|\psi_i(a)\| = \limsup\|\phi_{i,j}(a)\| = \lim \|\phi_{i,j}(a)\| $$ and the limit exists because the net $\{\|\phi_{i,j}(a)\|\}$ is decreasing. Now suppose $(B,\mu_j)$ is another system such that $$ \mu_j\circ\varphi_{i,j} = \mu_i $$ If $a\in A_i$, then $$ \|\mu_i(a)\| \leq \|\varphi_{i,j}(a)\| \quad\forall j\geq i $$ Hence, $\|\mu_i(a)\| \leq \|\psi_i(a)\|$. Hence, $\ker(\mu_i) \subset \ker(\psi_i)$, so there is a unique $\ast$-homomorphism $$ \widetilde{\mu_i} : \psi_i(A_i) \to B $$ such that $\widetilde{\mu_i}\circ \psi_i = \mu_i$. By uniqueness, if $j\geq i$, $\widetilde{\mu_{j}}\lvert_{\psi_i(A_i)} = \widetilde{\mu_i}$. Hence, we obtain a $\ast$-homomorphism $$ \widetilde{\mu} : \bigcup_i \psi_i(A_i) \to B $$ such that $\widetilde{\mu}\lvert_{\psi_i(A_i)} = \widetilde{\mu_i}$. Furthermore, each $\widetilde{\mu_i}$ is norm decreasing, so $\widetilde{\mu}$ is norm-decreasing. Hence it extends to a $\ast$-homomorphism $$ \widetilde{\mu} : A\to B $$ One can then check that $\widetilde{\mu}$ is the unique $\ast$-homomorphism such that $\widetilde{\mu}\circ\psi_i = \mu_i$.

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    Thanks, but I am still not quite sure that your element $(b_j)$ lies in $\sum_j A_j$ because given any $\epsilon  > 0$ I am not sure how to find a cofinite subset $J \subset I$ with $j \in J$ implies $\lVert b_j \rVert < \epsilon$. Maybe one should take $\sum_j A_j$ to be the elemets $(a_j)$ with $\lVert a_j \rVert \to 0$ by viewing $(\lVert a_j \rVert)_j$ as a net in $\mathbb R$ indexed by the directed set $I$ ?2017-01-18
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    Take $J = \{k \in I : k\geq j\}$2017-01-18
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    Why is that a cofinite set ? Indeed, I my main question was what happens when $I$ is an arbitrary index set. If for example $I = \mathbb Z$ with natural oder, every set of the form $\{ k : k \geq j \}$ is not cofinite.2017-01-18