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The Jacobson radical is $J(R)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals. Every proper right ideal $I$ lies in such an $L$: the union of a chain of proper right ideals containing $I$ is a right ideal and remains proper because it omits $1$. Thus Zorn lemma applies.

The right-module form of Nakayama lemma says that a finitely generated module $M$ with $MJ=M$ is zero. More generally, if $N$ is a submodule and $M=N+MJ$, then $M=N$. To prove the first statement, suppose $M\ne 0$ and take a generating set $m_1,\dots ,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ withThe unit criterion for the Jacobson radical makes $1-a_s$ a unit. Hence $m_s$ is generated by the other $m_i$, a contradiction. Apply this result to the quotient module $M/N$ to obtain the general form.

For an injective module endomorphism $f$ of an Artinian module $M$, the descending chain condition givesfor some $n$. Given $x\in M$, write $f^n(x)=f^{n+1}(y)$; injectivity of $f^n$ gives $x=f(y)$. Thus $f$ is surjective, as asserted by Artinian modules are co-Hopfian.

Now $R$ is a semilocal ring and $\overline{V}=V/VJ$ is a finitely generated module over the right Artinian ring $\overline{R}=R/J$. A finite direct sum of copies of $\overline{R}$, and hence its quotient module $\overline{V}$, is an Artinian module. The module endomorphism $\alpha$ induces $\overline{\alpha }$ on $\overline{V}$, and the given inverse-image condition says precisely that $\ker \overline{\alpha }=0$. The preceding argument makes $\overline{\alpha }$ surjective, soThe quotient module $W=V/\alpha (V)$ is finitely generated and satisfies $WJ=W$. Applying Nakayama lemma yields $W=0$. Therefore $\alpha$ is surjective, by surjectivity from injectivity modulo the Jacobson radical.