Inductive limits of $C^*$-algebras 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 $Q$.
 A: 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 $\widetilde{\mu}\circ\psi_i = \mu_i$.
