Indeed, you cannot have an elementary embedding $j:V\rightarrow M$ with critical point $\omega_1^V$. However, there are some subtleties to keep in mind:

• We could have an elementary embedding $j:W\rightarrow M$, where $M$ is an inner model of $V$, with $crit(j)=\omega_1^V$ - for instance, let $W\models$"$\kappa$ is measurable", and let $V$ be a forcing extension of $W$ in which $\kappa$ is collapsed to $\omega_1$. (In fact, we can do even better - see Joel's comment below.)

• Also, note that, while in ZFC having a $\kappa$-complete ultrafilter on $\kappa$ means that $\kappa$ is the critical point of an elementary embedding, this fails in ZF alone - so, while ZF+AD proves "$\omega_1$ is measurable," this does not mean that in ZF+AD there is an elementary embedding of $V$ into an inner model $M$ with critical point $\omega_1$.

May I ask what argument you have seen?

Indeed, you cannot have an elementary embedding $j:V\rightarrow M$ with critical point $\omega_1^V$. However, there are some subtleties to keep in mind:

• We could have an elementary embedding $j:W\rightarrow M$, where $W$ is an inner model of $V$, with $crit(j)=\omega_1^V$ - for instance, let $W\models$"$\kappa$ is measurable", and let $V$ be a forcing extension of $W$ in which $\kappa$ is collapsed to $\omega_1$. (In fact, we can do even better - see Joel's comment below.)

• Also, note that, while in ZFC having a $\kappa$-complete ultrafilter on $\kappa$ means that $\kappa$ is the critical point of an elementary embedding, this fails in ZF alone - so, while ZF+AD proves "$\omega_1$ is measurable," this does not mean that in ZF+AD there is an elementary embedding of $V$ into an inner model $M$ with critical point $\omega_1$.

May I ask what argument you have seen?

Indeed, you cannot have an elementary embedding $j:V\rightarrow V$ with critical point $\omega_1^V$. However, there are some subtleties to keep in mind:

• We could have an elementary embedding $j: M\rightarrow M$, where $M$ is an inner model of $V$, with $crit(j)=\omega_1^V$ - for instance, let $M\models$"$\kappa$ is measurable", and let $V$ be a forcing extension of $M$ in which $\kappa$ is collapsed to $\omega_1$.

• Also, note that, while in ZFC having a $\kappa$-complete ultrafilter on $\kappa$ means that $\kappa$ is the critical point of an elementary embedding, this fails in ZF alone - so, while ZF+AD proves "$\omega_1$ is measurable," this does not mean that in ZF+AD there is an elementary embedding of $V$ into $V$ with critical point $\omega_1$.

May I ask what argument you have seen?

Indeed, you cannot have an elementary embedding $j:V\rightarrow M$ with critical point $\omega_1^V$. However, there are some subtleties to keep in mind:

• We could have an elementary embedding $j:W\rightarrow M$, where $M$ is an inner model of $V$, with $crit(j)=\omega_1^V$ - for instance, let $W\models$"$\kappa$ is measurable", and let $V$ be a forcing extension of $W$ in which $\kappa$ is collapsed to $\omega_1$. (In fact, we can do even better - see Joel's comment below.)

• Also, note that, while in ZFC having a $\kappa$-complete ultrafilter on $\kappa$ means that $\kappa$ is the critical point of an elementary embedding, this fails in ZF alone - so, while ZF+AD proves "$\omega_1$ is measurable," this does not mean that in ZF+AD there is an elementary embedding of $V$ into an inner model $M$ with critical point $\omega_1$.

May I ask what argument you have seen?