Why do the mapping from $A^m$ to $M_n$ is surjective? I was unable to solve the problem 1.3.11(b) from Qing Liu's Algebraic Geometry and Arithmetic Curves.
Let $A$ be a commutative complete ring for the $I$-adic topology with unit, where $I$ is an ideal of $A$. Let $(M_n)_{n\geq 0}$ be $A$-modules such that $I^{n+1}M_n=0$ 
and that there exist a surjective homomorphisms $\pi_n:M_{n+1}\to M_n$ with 
$\operatorname{Ker} (\pi_n)=I^{n+1}M_{n+1}$. Let $M=\varprojlim_n M_n$ and denote the (surjective) canonical homomorphism by $u_n:M\to M_n$.
Suppose that $M_0$ is generated over $A$ by a finitely number of elements $e_{0,1},\ldots, e_{0,m}$. Let $e_1,\ldots,e_m\in M$ be such that $u_0(e_i)=e_{0,i}$, and define $\phi_n:A^m\to M_n$ by $(a_1,\ldots,a_m)\mapsto \sum_ia_iu_n(e_i)$. How can I show that $\phi_n$ is surjective and $M$ is generated by the $e_i$? Do I need some kind of universal property?
 A: $M_n\require{AMScd}$ is finitely generated

We have an exact sequence
\begin{CD}
  0 @>>> IM_n @>{\hookrightarrow}>> M_n @>{\pi_{0n}}>> M_0 @>>> 0
\end{CD}
where $\pi_{0n} = \pi_0 \circ \cdots \circ \pi_{n-1}$ (see (a) of this exercise).
Let $m_1, m_2, \ldots \in M_n$, and $M_n$ is generated by $u_n(e_1), \ldots, u_n(e_m), m_1, m_2, \ldots$ over $A$:
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A + \sum_{k\geq1} m_k A . \]



Let us consider $\pi_{0n}(m_1) \in M_0$. Since $M_0$ is generated by $u_0(e_1), \ldots, u_0(e_m)$, there exist $a_1, \ldots, a_m \in A$ such that
\begin{align}
  \pi_{0n}(m_1) &= a_1 u_0(e_1) + \cdots + a_m u_0(e_m) \\ 
  &= \pi_0( a_1 u_1(e_1) + \cdots + a_m u_1(e_m) ) \\
  &= \pi_0 \circ \cdots \circ \pi_{n-1} ( a_1 u_n(e_1) + \cdots + a_m u_n(e_m) ) \\
  &= \pi_{0n}(a_1 u_n(e_1) + \cdots + a_m u_n(e_m)).
\end{align}
Since $\ker \pi_{0n} = IM_n$, and $M_n$ is generated by $u_1, \ldots, u_m$ and $m_k~(k\geq1)$, there exits $i_1, \ldots, i_m \in I$ and $\tilde{i}_k~(k\geq1) \in I$ such that
\[ m_1 - (a_1 u_n(e_1) + \cdots + a_m u_n(e_m)) = i_1 u_n(e_1) + \cdots i_m u_n(e_m) + \sum_{k\geq1} \tilde{i}_k m_k. \]
Hence
\[  m_1 = (a_1 + i_1) u_n(e_1) + \cdots + (a_m + i_m) u_n(e_m) + \sum_{k\geq2} \tilde{i}_k m_k. \]
Since $A$ is complete respect to $I$-adic topology, $1 - \tilde{i}_1$ is invertible in $A$ (in fact, $1 - \tilde{i}_1 + \tilde{i}_1{}^2 - \cdots$ is its inverse). Hence we have
\[ m_1 = (1 - \tilde{i}_1)^{-1}\sum_{l=1}^m(a_l + i_l) u_n(e_l)+ \sum_{k\geq2} (1 - \tilde{i}_1)^{-1}\tilde{i}_k m_k. \]
Consequently, we can express $M_n$ as
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A + \sum_{k\geq2} m_k A. \]


In the same manner, we can show that $M_n$ is generated by $u_n(e_1), \ldots, u_n(e_m)$:
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A . \]
In other words, 
\[ \phi_n \colon A^{\oplus m} \ni (a_1, \ldots, a_m) \mapsto \sum_{k=1}^m a_k u_n(e_k) \in M_n \]
is surjective.

$M$ is finitely generated
I just give a brief summary of proof. There would be much better ways.

Since $I^{n+1}M_n = 0$, we have a surjection
\[ \bar{\phi}_n \colon (A/I^{n+1})^{\oplus m} \ni ([a_1], \ldots ,[a_m]) \mapsto \sum_{k=1}^m a_k u_n(e_k) \in M_n . \]
Consider the commutative diagram
\begin{CD}
  0 @>>> \ker \bar{\phi}_{n+1} @>{\hookrightarrow}>> (A/I^{n+2})^{\oplus m} @>{\bar{\phi}_{n+1}}>> M_{n+1} @>>> 0 \\
  @. @VV{p_n'}V @VV{p_n}V @VV{\pi_n}V @. \\
  0 @>>> \ker \bar{\phi}_{n} @>{\hookrightarrow}>> (A/I^{n+1})^{\oplus m} @>{\bar{\phi}_{n}}>> M_{n} @>>> 0
\end{CD}
where $p_n \colon (A/I^{n+2})^{\oplus m} \to (A/I^{n+1})^{\oplus m}$ is given by
\[ p_n \colon (A/I^{n+2})^{\oplus m} \ni ([a_1], \ldots , [a_m]) \mapsto ([a_1], \ldots , [a_m]) \in (A/I^{n+1})^{\oplus m} . \]
Since $p_n'$ is surjective, by Lemma in the book, we have an exact sequence
\begin{CD}
  0 @>>> \varprojlim (\ker \bar{\phi}_{n}) @>>> \bigoplus_{k=1}^m \varprojlim (A/I^{n+1}) @>{\phi}>> \varprojlim M_n @>>> 0
\end{CD}
, which immediately follows ($A \simeq \varprojlim A/I^{n+1}$)
\begin{CD}
  A^{\oplus m} @>{\phi}>> M @>>> 0
\end{CD}
$\phi$ is given by
\[ \phi \colon A^{\oplus m} \ni (a_1, \ldots, a_m) \mapsto \sum_{k=1}^m a_k e_k \in M. \]
