= Solution
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 $M\ne0$ and take a generating set $m_1,\ldots,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ with
$$
m_s=\sum_{i=1}^s m_i a_i,\qquad
m_s(1-a_s)=\sum_{i<s}m_i a_i.
$$
The <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> gives
$$
f^n(M)=f^{n+1}(M)
$$
for 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>, so
$$
V=\alpha(V)+VJ.
$$
The <quotient module> $W=V/\alpha(V)$ is finitely generated and satisfies $WJ=W$. Applying <Nakayama lemma> yields $W=0$. Therefore \b[$\alpha$ is surjective], by <surjectivity from injectivity modulo the Jacobson radical>.
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