Suppose $R \sim_\omega R'$. Then for every $k$-tuple $a$ in $E$ and every natural number $p$, there is a $k$-tuple $b$ in $E'$ such that $a \sim_p b$ Sorry to bother you guys again with a Poizat question, but I'm struggling a little bit with the material (as it must be obvious) and I want to check if I got the main idea correctly or if I'm totally off. I'll give the basic definitions again and proceed to my question, which is basically about Exercise 1.1 from Poizat's A Course in Model Theory (sorry for the repition, but it's my understanding that questions in this site must be self-contained).
Poizat defines an isomorphism $s$ from a relation $R$ to a relation $R'$ as a bijection from $E$ to $E'$ (where these denote the respective underlying sets, or universes) which preserves the given relation (i.e. a sequence $\vec{a}$  satisfies $R$ iff the sequence $s(\vec{a})$ also satisfies $R'$). After defining the notion of a local isomorphism from $R$ to $R'$ as an isomorphism between finite restrictions of $R$ to $R'$ (where a finite restriction is a restriction of the relation to a finite subset of the original universe), he proceeds to define the set $S_p(R, R')$ of $p$-isomorphisms between $R$ and $R'$ ($p$ is a non-negative integer). The definition is recursive: $S_0(R,R')$ is defined as the set of all local isomorphisms between $R$ and $R'$ and, for $p>0$, we say that a local isomorphism $s$ belongs to the set $S_{p+1}(R,R')$ according to the following conditions:
Back: For every $a$ in the universe $E$ of $R$, there is an extension $t$ of $s$ (i.e., $\mathrm{dom}(s)\subseteq \mathrm{dom}(t)$ and $s$ is the restriction of $t$ to $\mathrm{dom}(s)$), defined at $a$, that is in $S_p(R,R')$;
Forth: For every $b$ in the universe $E'$ of $R'$, there is an extension $t$ of $s$, whose image contains $b$, that is in $S_p(R,R')$.
By an $\omega$-isomorphism, we mean a local isomorphism $\sigma$ from $R$ to $R'$ that is a $p$-isomorphism for every nonnegative integer $p$. We write $S_\omega(R, R')$ for the set of all $\omega$-isomorphisms between $R$ and $R'$, i.e. the intersection of all $S_p(R, R')$. If $S_\omega(R, R')$ is non-empty, we say that $R$ and $R'$ are elementary equivalent, in symbols, $R \sim_\omega R'$. If $f_\emptyset$ (the empty function) is a $p$-isomorphism from $R$ to $R'$, we say that they are $p$-equivalent, and denote this by $R \sim_p R'$. If there is a $p$-isomorphism from $R$ to $R'$ according to which a given $k$-tuple $\vec{a} = \langle a_1, \dots, a_k \rangle$ in the universe of $R$ corresponds to a $k$-tuple $\vec{b}=\langle b_1, \dots, b_k \rangle$ in the universe of $R'$, we say that these $\vec{a}$ and $\vec{b}$ are $p$-equivalent. 
Given this, in Exercise 1.1 (p. 4), Poizat asks us to prove the following. Suppose $R$ and $R'$ are elementary equivalent. We claim that, for every $k$-tuple $\vec{a}$ in the universe of $R$ and every nonnegative integer $p$, there is a $k$-tuple $\vec{b}$ in the universe of $R'$ such that $(\vec{a}, R) \sim_p (\vec{b}, R')$.
Now, here's what I thought. The proof is by induction on$k$. For $k = 0$, since $R$ and $R'$ are elementary equivalent, $f_\emptyset$ will be in every $S_p(R, R')$, so that takes care of it. For $k=n+1$, Suppose the  hypothesis of the theorem and consider an arbitrary $k$-tuple $\vec{a}_k$  and an arbitrary $p$. By the induction hypothesis, for every $n$-tuple and for every nonnegative integer $i$, there is an $n$-tuple $\vec{b}$ such that $(\vec{a}, R) \sim_i (\vec{b}, R')$. In particular, for every $n$-tuple, there is a $p+1$-isomorphism $\sigma$ from $R$ to $R'$ such that, for each $j \leq n$, $a_j \in \operatorname{dom}(\sigma)$. We can thus select an $n$-tuple $\vec{a}_n$ such that, for each $a_j \leq n$, $\vec{a}_n$ agrees with $\vec{a}_k$ and such that there is a corresponding $p+1$- isomorphism, say $\tau$, from $R$ to $R'$. Now, by the forth condition, it follows that, for every $a \in E$, there's an extension $\tau'$ of $\tau$, such that $a \in \operatorname{dom}(\tau')$ and $\tau' \in S_p(R, R')$. In particular, there's an extension $\tau'$ of $\tau$ such that $a_k \in \operatorname{dom}(\tau')$. Thus, for each $j \leq k$, $a_j \in \operatorname{dom}(\tau')$. Now, consider $\tau'(\vec{a}_k)$. This will be a $k$-tuple $\vec{b}$ in the domain of $R'$, thus completing the proof.
Is the above reasoning correct? I'm particularly concerned with the last step, namely taking $\tau'(\vec{a}_k)$. It seems that, given $\tau'$, it's possible to define the action of $\tau'$ on the tuple $\vec{a}_k$ by describing the action of $\tau'$ on each element of $\vec{a}_k$). I'm not sure, however, that this is legitimate...
 A: Well, since nobody has answered, I'll post the answer which I have worked out.
Upon reflection, I believe that it's  not necessary to proceed by induction. One can argue in this way:
Suppose the hypothesis of the theorem and consider an arbitrary $k$-tuple $\vec{a}_k$ and an arbitrary $p$. By the hypothesis of the theorem, we know that $f_\emptyset$ is a $n$-isomorphism for every nonnegative integer $n$. In particular, it's $p+k$-isomorphism. Now, by the forth condition, it follows that, for every $a \in E$, there's an extension $\tau$ of $f_\emptyset$, such that $a \in \operatorname{dom}(\tau)$ and $\tau \in S_{p+(k-1)}(R, R')$. We can therefore select a $\tau$ such that $a_1 \in \operatorname{dom}(\tau)$. Considering again the forth condition, there is an extension $\tau'$ of $\tau$ such that $\tau'$ is in $S_{p+(k-2)}(R, R')$ such that $a_2 \in \operatorname{dom}(\tau')$. Repeating this procedure $k$-times, we eventually arrive at a $p$-isomorphism $\sigma$ such that each $a_i$ ($i \leq k$) is in $\operatorname{dom}(\sigma)$. Thus, for each $a_i$, $\sigma(a_i) = b_i$ for some $b_i \in E'$. Therefore, the tuple $\vec{b} = \langle b_1, \dots, b_k \rangle$ is in $E'^k$, that is, it's in the universe of $R'$.
In the original question, I was a bit confused with the expression "the universe of $R'$"; I thought the $k$-tuple $\vec{b}$ needed to be in $E'$ (sort of), and then there was no guarantee that the $p$-isomorphism would take the tuple $\vec{a}$ to another tuple. Now I see that the only requirement is that $\vec{b}$ needs to be an element of $E'^k$, which follows clearly from the above (plus some set-theoretical assumptions, I guess).
EDIT: Thinking a bit more about the tuple issue, it seems to me that the idea is a bit more involved. Poizat does not mention that the $k$-tuple needs to be an element of $E^k$; rather, he appears to be thinking of the tuple as a function from $k$ to $E$. Thus, in order to generate the $k$-tuple $\vec{b}$, I need to index each $b$ according to the corresponding $a_i \in \vec{a}$. For instance, if the first extension $p+(k-1)$-isomorphism $\tau$ is such that $\tau(a_1) = b$, then I index $b$ as $b_1$. The next extension will give me $\tau'(a_2) = b'$, so I set $b'=b_2$, etc. This way, $a_i = a_j$ iff $b_i = b_j$ and the $p$-isomorphism $\sigma$ will be $\sigma(a_n) = b_n$ for $n \leq k$.
A: I will add a long comment regarding the tuple issue.
Poizat's definition of universe [page 1] is "standard" : 

an $m$-ary relation $R$ with universe $E$ is a subset of $E^m$.

Thus, at the very beginning of his treatise, he introduce the notion of $m$-tuple :

if the $m$-tuple $\overline a = (a_1,\ldots, a_m)$ in [the universe] $E$ belongs to $R$, we say that it satisfies the relation $R$.

[see the original french version, page 14 : 


si le $m$-uple $\overline a$ extrait de $E$ appartient à $R$ on dit qu'il satisfait la relation $R$].


It seems to me that it is only a "shorthand" , and we have to read : 

the $m$-tuple $\overline a$ in $E$ 

as : 


the $m$-tuple $\overline a$ [is a finite sequence] of elements belonging to $E$.


The same issue we can find in page 19, when Poizat speaks of (first-order) formulae :

[consider] a formula in the form $f(\overline x)$, where $\overline x$ is a $n$-tuple of variables $(x_1,\ldots, x_n)$. [...] Let us consider an $m$-ary relation $R$, a formula $f(\overline x)$, and an $n$-tuple $\overline a = (a_1,\ldots, a_n)$ in the universe of $R$ [...].


There is also another way to look at a finite sequence $(a_1,\ldots, a_n)$ of $n$ elements in $A$.
We can consider it a a set of pairs : $\{ (1,a_1),\ldots, (n,a_n) \}$. This is a function defined on the set $\{ 1,\ldots, n \}$, i.e. :

$f : n_0^+ \to A$ [where $n_0 = n \backslash \{ 0 \}$ and $n^+ = n \cup \{ n \}$].

