Inverse limit: show the surjectivity of $\mu_{ij}$ Let $A_i$ rings and $$P=\left\{(a_i)\in \prod_i A_i\mid\forall i\leq j, \mu_{ji}(a_j)=a_i\right\}$$
where $\mu_{ji}:A_j\to A_i$ are rings homomorphism. We defined $\mu_i: P\to A_i$ by the projection on the $i$-th component. We want to prove that if $\mu_{ji}$ are surjective (for all $i$ and $j$), then so is every $\mu_i$. There is a proof here (part 2.). I don't understand why they introduce $t$ in the second part of the proof. 
I mean, when they write :
For base case, let $j=i=0$. Then $\mu_{j,i}(a_j)=a_i$. Suppose now for some $t\geq 0$ whenever $i\leq j\leq t$, $\mu_{ji}(a_j)=a_i$. Now consider...
Why do they introduce this $t$? It doesn't look really necessary, no? Any explanation would be more than appreciated.
 A: Edit: I have changed my answer, showing that $t$ is actually unnecessary.
The use of the variable $t$ is indeed superfluous, and the author did not make it transparent why this variable appears at all. Let us step through the proof of part 2, rewriting the proof by induction at the end so as to not use $t$.
We are given rings 
$$A_0,A_1,A_2,\dots,$$
and ring homomorphisms that go from higher indices to lower indices:
$$\mu_{ji}\colon A_j\to A_i\quad\text{where $i\le j$}.$$
The ring homomorphisms are assumed to satisfy two conditions: First, $\mu_{ii}$ is the identity homomorphism from $A_i$ into $A_i$ for all $i$. Second, the following composition rule holds:
$$\mu_{ji}\circ\mu_{kj}=\mu_{ki}\quad\text{for all $i\le j\le k$}.$$
The set you call $P$ is the set of all tuples $(a_i)$ in the product $\prod_{i\ge 0}A_i$ that are in a sense closed under the homomorphisms $\mu_{ji}$. More precisely,
$$\mu_{ji}(a_j)=a_i\quad\text{for all $i\le j$}.$$
Now, the claim is that when the homomorphisms $\mu_{ji}$ are all surjective, then for any $k\in\mathbb N$ and any $a\in A_k$, there exists $(a_i)\in P$ such that $a_k=a$. In other words, the $k$th coordinate projection of $P$ is surjective.
Let $k\in\mathbb N$ and $a\in A_k$. The author constructs a tuple
$$(a_i)\in\prod_{i\in\mathbb N} A_i$$
such that $a_k=a$, as follows. If $i\le k$, we can simply use our homomorphism $\mu_{ki}$, defining
$$a_i=\mu_{ki}(a).$$
If $i>k$ and $a_{i'}$ has been defined for all $i'<i$, then we use the surjective property of $\mu_{i,i-1}$ and the given $a_{i-1}$. Indeed, since $\mu_{i,i-1}$ is surjective and $a_{i-1}\in A_{i-1}$, there exists some $x\in A_i$ such that
$$\mu_{i,i-1}(x)=a_{i-1}.$$
Thus, we define $a_i=x$. Recursively, $a_i$ is defined for all $i\in\mathbb N$.
Now, we need to show that $(a_i)$ belongs to $P$. It is evident from the construction that $a_i\in A_i$ for all $i\in\mathbb N$. We need to prove $\mu_{ji}(a_j)=a_i$ for all natural numbers $i\le j$. We will prove that by induction on $j$.
The basis step: $j=0$. Let $i\le j$, so that $i=0$ as well. Then, $\mu_{ji}=\mu_{00}$ is the identity on $A_0$, so $\mu_{ji}(a_j)=a_j=a_0$ and $a_i=a_0$, implying $\mu_{ji}(a_j)=a_i$ in this case.
The induction step: assume that $j$ is a natural number such that $\mu_{ji}(a_j)=a_i$ for all $i\le j$. Let $i\le j+1$. Either $i=j+1$ or $i\le j$. In the first case, $\mu_{j+1,i}(a_{j+1})=a_i$ since $\mu_{j+1,i}$ is the identity and $a_i=a_{j+1}$. We will now assume $i\le j$. We will first show that
$$\mu_{j+1,j}(a_{j+1})=a_j$$
using a proof by cases.
If $j+1\le k$, then
$$a_{j+1}=\mu_{k,j+1}(a)\quad\text{and}\quad a_j=\mu_{kj}(a)$$
so 
$$\mu_{j+1,j}(a_{j+1})=(\mu_{j+1,j}\circ\mu_{k,j+1})(a)=\mu_{kj}(a)=a_j$$
If $j>k$, then as we constructed $(a_i)$ above,
$$\mu_{j+1,j}(a_{j+1})=a_j.$$
Therefore, $\mu_{j+1,j}(a_{j+1})=a_j$. Furthermore, by the induction hypothesis, and the assumption $i\le j$,
$$\mu_{ji}(a_j)=a_i.$$
Then,
$$\mu_{j+1,i}(a_{j+1})=(\mu_{ji}\circ\mu_{j+1,j})(a_{j+1})=\mu_{ji}(a_j)=a_i.$$
By our proof by induction we have shown that $\mu_{ji}(a_j)=a_i$ for all $i\le j$. Therefore, $(a_i)\in P$.
This revision of the author's proof illustrates that there is no need to make use of an additional variable $t$ for the induction.
