Trying to understand a proof from math overflow regarding direct limit 
Direct limit of $\mathbb {C^*}$ behaves well with quotients.

The following solution is from mathoverflow:
Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $\phi_n: A_n \to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $\phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $\tilde{\phi}_n: A_n/I_n \to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$.
This is easy because the maps $\tilde{\phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.

I understand the whole argument whatever is written but I don’t follow how does this proves that $A/I$ is direct limit? Please explain what are we using here?

 A: In general, if $A_1\xrightarrow{\phi_1}A_2\to\dots$ is an inductive sequence and $(A, \{\mu_n\})$ is a $C^*$-algebra with $*$-homomorphisms $\mu_n:A_n\to A$ so that $\mu_{n+1}\circ\phi_n=\mu_n$ for all $n$, then $(A,\{\mu_n\})$ is the inductive limit of the sequence if and only if

*

*$A=\overline{\bigcup_{k=1}^\infty\mu_k(A_k)}$

*$\ker(\mu_n)=\{x\in A_n: \lim_{m\to\infty}\|\phi_{m,n}(x)\|=0\}$ for all $n$, where $\phi_{m,n}=\phi_{m-1}\circ\phi_{m-2}\circ\dots\circ\phi_n$.

Now if $I$ is a (closed, two-sided) ideal in $A$, then set $I_n:=\mu_n^{-1}(I)$. Note that $I_n$ is a closed, two-sided ideal in $A_n$. We now set $\psi_n:A_n/I_n\to A_{n+1}/I_{n+1}$ by $\psi_n(x+I_n)=\phi_n(x)+I_{n+1}$. These are well-defined, since if $\phi_n(x)-\phi_n(x')\in I_{n+1}$ then $\mu_{n+1}(\phi_n(x-x'))\in I$, so $\mu_n(x-x')\in I$ so $x-x'\in I_n$. It is easy to see that each $\psi_n$ is a *-homomorphism. Moreover, if $\psi_n(x+I_n)=0$, then $\varphi_n(x)\in I_{n+1}$, so $\mu_{n+1}(\varphi_n(x))\in I$, so $\mu_n(x)\in I$ so $x\in I_n$, so $x+I_n=0$, thus the maps $\psi_n$ are injective. So we have an inductive sequence
$$A_1/I_1\xrightarrow{\psi_1}A_2/I_2\to\dots $$
where the maps $\psi_n$ are injective. Note that if we set $\pi_n:A_n/I_n\to A/I$ as $\pi_n(x+I_n)=\mu_n(x)+I$, then the $\pi_n$'s are well-defined, injective $*$-homomorphisms (verified as we did for $\psi_n$). Moreover,
$$\pi_{n+1}\circ\psi_n(x+I_n)=\pi_{n+1}(\phi_n(x)+I_{n+1})=\mu_{n+1}(\phi_n(x))+I=\mu_n(x)+I=\pi_n(x+I_n)$$
so $\pi_{n+1}\circ\psi_n=\pi_n$. If we verify the conditions (1) and (2) for the system of quotiens, then we can conclude that $\varinjlim(A_n/I_n, \psi_n)=(A/I,\{\pi_n\})$. But indeed, (2) is trivial since $\ker(\pi_n)=0$ and $\|\psi_{m,n}(x)\|=\|x\|$ for all $m$, since each $\psi_k$ is injective (thus isometric) so $\{x\in A_n: \lim_{m\to\infty}\|\psi_{m,n}(x)\|=0\}=\{0\}$.
Condition (1) is easily verified: let $x+I\in A/I$, $\varepsilon>0$. Since $x\in A$, we may find $k\geq1$ and $y\in A_k$ so that $\|x-\mu_k(y)\|<\varepsilon$, because $A=\overline{\bigcup_{k=1}^\infty\mu_k(A_k)}$ (since $A$ is the inductive limit of $A_n$). But now
$$\|(x+I)-\pi_k(y+I_k)\|=\|(x+I)-(\mu_k(y)+I)\|=\|(x-\mu_k(y))+I\|\leq\|x-\mu_k(y)\|<\varepsilon $$
and we are done. Does this answer your question?
