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Yes, it is sufficient. Your question actually touches on a more general problemknown, similar issue. A proper class, generally, is not an object in ZF, so in ZF we can't directly refer to it. Hence, again under the assumption we work in ZF, $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinationsDespite all this, how can we refer to thosemathematicians working in ZF talk about these things? indirectly.

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete partswithout accepting the existence of $V$. The method is the same as the one outlined above.

Another way of looking at your problem is to think of a less complicated, thatsimilar scenario. In classical potentialism, one refutes the existence of $\mathbb{N}$ but accepts all its finite initial segments. Here, the problem would be whether one can talk about functions $f \colon \mathbb{N} \to A$, where $A$ is in termssome finite set of natural numbers. In particular, let us call $V_\alpha$$n$ a pigeonhole number whenever there is an $m<n$ such that $f(m) = f(n)$.

The question: For every finite set of natural numbers $A$, does every function $f \colon \mathbb{N} \to A$ admit a pigeonhole number? While $f$ and $\mathbb{N}$ don't exist for a classical potentialist, he can still answer this question! He can't do it explicitly, but he can do it implicitly using larger and larger finite initial segments of $\mathbb{N}$.

Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

Yes, it is sufficient. Your question actually touches on a known, similar issue. A proper class is not an object in ZF, so in ZF we can't directly refer to it. Hence, again under the assumption we work in ZF, $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. Despite all this, mathematicians working in ZF talk about these things indirectly.

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ without accepting the existence of $V$. The method is the same as the one outlined above.

Another way of looking at your problem is to think of a less complicated, similar scenario. In classical potentialism, one refutes the existence of $\mathbb{N}$ but accepts all its finite initial segments. Here, the problem would be whether one can talk about functions $f \colon \mathbb{N} \to A$, where $A$ is some finite set of natural numbers. In particular, let us call $n$ a pigeonhole number whenever there is an $m<n$ such that $f(m) = f(n)$.

The question: For every finite set of natural numbers $A$, does every function $f \colon \mathbb{N} \to A$ admit a pigeonhole number? While $f$ and $\mathbb{N}$ don't exist for a classical potentialist, he can still answer this question! He can't do it explicitly, but he can do it implicitly using larger and larger finite initial segments of $\mathbb{N}$.

moved exceptional case to where it should be in the first place
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Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding, with exception to the case where $j \colon V \to V$, is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can assume that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding, with exception to the case where $j \colon V \to V$, is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

added 2 characters in body
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Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can showassume that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding, with exception to the case where $j \colon V \to V$, is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

Yes, it is sufficient. Your question actually touches on a more general problem. A class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can show that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding, with exception to the case where $j \colon V \to V$, is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

Yes, it is sufficient. Your question actually touches on a more general problem. A proper class, generally, is not an object. Hence $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. So, regardless of our philosophical inclinations, how can we refer to those things?

The trick is, one can assume that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding, with exception to the case where $j \colon V \to V$, is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ in terms of incomplete parts of $V$, that is in terms of $V_\alpha$.

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