I read the following solution for Showing that NP is closed under union and they used the same $c$ for both the verifies $V_1$ and $V_2$. Why is it correct?
Let $L_1$ and $L_2$ be languages in $NP$. Also, for $i = 1, 2$ let $V_i(x, c)$ be an algorithm that, for a string $x$ and a possible certificate $c$, verifies whether $c$ is actually a certificate for $x \in L_i$. Thus, $V_i(x,c) = 1$ if certificate c verifies $x \in L_i$, and $V_i(x, c) = 0$ otherwise. Since both $L_1$ and $L_2$ are both in $NP$, we know that $V_i(x, c)$ terminates in polynomial time $O(|x|^d)$ for some constant $d$. To show that $L_3 = L_1 \cup L_2$ is also in $NP$, we will construct a polynomial-time verifier $V3$ for $L_3$. Since a certificate $c$ for $L_3$ will have the property that either $V_1(x, c) = 1$ or $V_2(x, c) = 1$, we can easily construct a verifier $V_3(x, c) = V_1(x, c) \lor V_2(x, c)$. Clearly then $x \in L_3$ if and only if there is a certificate $c$ such that $V_3(x, c) = 1$. Notice also that the new verifier $V_3$ will run in time $O(2(|x|^d))$, which is polynomial. Therefore, the union $L_3$ of two languages in $NP$ is also in $NP$, so $NP$ is closed under union.
taken from here.
$M_1,M_2$ TM which accept $w$ can accept $w$ for different reasons so we can't claim that $c_1=c_2$
Is it legal to use the same certificate for both $V_1,V_2$ in the answer? Why?
Is a verifier by definition is deterministic TM?
In the above answer, does $V3$ runs $x,c$ on both $V_1$ and $V_2$ in the worst case?