= Solution
Write $h_\alpha$ for the simple <coroot> and choose the corresponding normalized $\mathfrak{sl}_2$ triple. The order condition means
$$
\chi-s_\alpha\chi=k\alpha\in Q_+,\qquad
k=\langle\chi,\alpha^\vee\rangle\in\mathbb Z_{\ge0}.
$$
The highest vector has $h_\alpha$-<representation weight> $\langle\chi-\rho,\alpha^\vee\rangle=k-1$. The raising-lowering identity gives
$$
E_\alpha E_{-\alpha}^km_\chi
=k((k-1)-k+1)E_{-\alpha}^{k-1}m_\chi=0.
$$
Other simple <positive root> vectors commute with $E_{-\alpha}$ and also kill this vector. Thus it is a highest vector of <representation weight>
$$
\chi-\rho-k\alpha=s_\alpha\chi-\rho.
$$
The universal property of the <Verma module> gives the required homomorphism. It is injective: under the <Poincaré-Birkhoff-Witt theorem> identification of each <Verma module> with the free rank-one $U(\mathfrak n_-)$ <module>, it is right multiplication by $E_{-\alpha}^k$. The algebra $U(\mathfrak n_-)$ has no zero divisors, because its associated graded algebra is the <polynomial> domain $S(\mathfrak n_-)$. Hence
$$
\boxed{M^{s_\alpha\chi}\hookrightarrow M^\chi,\qquad
m_{s_\alpha\chi}\mapsto E_{-\alpha}^km_\chi.}
$$
For $k=0$ this is the identity map, so the equality case of the order is covered.
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