We have to review the basic semantical notion for first-order logic.
See Herbert Enderton, A Mathematical Introduction to Logic, (2nd - 2001), SECT 2.2 : Truth and Models [page 80-on] :
Let $\varphi$ a first-order formula of the language $\mathcal L$, $\mathcal M$ a structure for the language and
$s : Var \rightarrow |\mathcal M|$ a function from the set $Var$ of all variables of the language ($v_0, v_1, v_2, ...$) into the universe (or domain) $|\mathcal M|$ of $\mathcal M$.
The function $s$ is often called assignment function, because it "assign" objects of the domain of interpretation to variables of the language.
Then we
we will define what it means for $\mathcal M$ to satisfy $\varphi$ with $s : Var \rightarrow |\mathcal M|$; in symbols : $\mathcal M \vDash \varphi[s]$.
We have to state the definition for terms and then for atomic formulae.
Basically, for an $n$-place predicate parameter $P$ (a predicate letter), we have that :
$\mathcal M \vDash P(t_1, ... t_n)[s]$ iff $(s(t_1), ... s(t_n)) \in P^{\mathcal M}$,
where the $n$-ary relation $P^{\mathcal M} \subseteq |\mathcal M|^n$ is is a set of $n$-tuples of members of the universe which "interpret" the symbol $P$.
Then we have the condition for the quantifiers :
$\mathcal M \vDash \forall x \varphi[s]$ iff for all $d \in |\mathcal M|$, we have $\mathcal M \vDash \varphi[s(x/d)]$, where $s(x/d)$ is the function which is exactly like $s$ except for one thing: at the variable $x$ it assumes the value $d$.
$\mathcal M \vDash \exists x \varphi[s]$ iff for some $d \in |\mathcal M|$, we have $\mathcal M \vDash \varphi[s(x/d)]$, where $s(x/d)$ is the function which is exactly like $s$ except for one thing: at the variable $x$ it assumes the value $d$.
Finally, we have that [page 88] :
$\varphi$ is valid iff for every $\mathcal M$ and every $s : Var \rightarrow |\mathcal M|$, $\mathcal M$ satisfies $\varphi$ with $s$.
Going back to the function $s$, consider a language $\mathcal L_{Ar}$ for arithmetic, its interpretation $\mathbb N$ and a formula $\varphi$ with one free variable :
$v_0 = 0$.
In order to "give meaning" to the formula, we have to "give a reference" to the variable $v_0$; we do this through an assignment $s : Var \rightarrow \mathbb N$ and let (for example) $s(v_0) = 0$.
Clearly $s$ satisfy $\varphi$, i.e $\mathbb N \vDash \varphi[s]$, because :
$(v_0 = 0)[s]$ is $0 = 0$, which is true in $\mathbb N$.
If we choose a "new" $s'$ with $s'(v_0)=1$, then $\mathbb N \nvDash \varphi[s']$, because $(v_0 = 0)[s']$ is $1 = 0$, which is false.
Now for the quantifiers.
Why :
For sentences, i.e. 'closed formulas', formulas without free variables, the mapping does not matter ?
We have two cases :
(i) $\varphi$ is a formula without quantifiers; being "closed" it must be a truth-functional composition of "closed" atomic formulae, like (in the above language for $\mathbb N$) :
$0 = 1$.
We have no free variables in this formula; thus we have no need to use an assignment $s$ to "give reference" to them.
The formula in this example is false, and we can say that is so for every assignement.
Consider now the other case :
(ii) $\varphi$ is $\forall v_0 \psi(v_0)$;
being closed, $v_0$ is the only variable free in $\psi$.
We apply the above clause for the semantics of quantifiers and we have that a function $s$ satisfy $\varphi$ iff all functions $s'$ which are exactly like $s$ except for the value at the variable $v_0$ satisfy $\psi$.
Again, if we consider our formula $\forall v_0(v_0 = 0)$ we have that it is (as expected) false according to the above definition.
Consider $s$ such that $s(v_0)=0$; clearly $s(v_0/1)$, i.e. the function which is exactly like $s$ except for one thing: at the variable $v_0$ it assumes the value $1$, does not satisfy $(v_0 = 0)$, i.e. $\mathbb N \nvDash (v_0 = 0)[s(x/1)]$.
Thus, we can conclude that $\mathbb N \nvDash ∀v_0 (v_0 = 0)[s]$, i.e. $∀v_0 (v_0 = 0)$ is not true in $\mathbb N$.
The same applies for the existential quantifier.
The above argument can be generalized in order to get [see page 86] :
Theorem 22A : Assume that $s_1$ and $s_2$ are functions from $Var$ into $|\mathcal M|$ which agree at all variables (if any) that occur free in the wff $\varphi$. Then
$\mathcal M \vDash \varphi[s_1]$ iff $\mathcal M \vDash \varphi[s_2]$.
Proof. Because satisfaction was defined recursively, this proof uses induction. We consider the fixed structure $\mathcal M$ and show by induction that every wff $\varphi$ has the property that whenever two functions $s_1, s_2$ agree on the variables free in $\varphi$ then $\mathcal M$ satisfies $\varphi$ with $s_1$ iff it does so with $s_2$.
Finally [page 87] :
Corollary 22В : For a sentence $\sigma$, either :
(a) $\mathcal M$ satisfies $\sigma$ with every function $s$ from $Var$ into $|\mathcal M|$, or
(b) $\mathcal M$ does not satisfy $\sigma$ with any such function.
