Finite-dimensional tensoring reaches every simple of a linkage class

Choose the finite bound. Since a simple reflection reverses only ${\alpha }_i$ among the positive roots, their half-sum satisfies $\rho -s_i\rho ={\alpha }_i$, giving the simple coroot pairings one. Thus $\rho$ is regular dominant integral. By [F1], $d_w=\rho -w\rho \in Q^{+}\setminus \{0\}$ for $w\mathrm{\ne }1$; take the first positive coordinate in the fixed finite simple-root enumeration. The finitely many real numbers $-\mathrm{Re}(c_w{)}_{j(w)}\mathrm{/}(d_w{)}_{j(w)}$ have an upper bound, so choose an integer $N\ge 0$ strictly larger than all of them. For rank zero $W=\{1\}$ and there are no restrictions; use $N=0$. With $\lambda =\mu +N\rho$, the identity $\lambda -w\cdot \lambda =c_w+N{d}_w$ has positive real part at the selected coordinate for each $w\mathrm{\ne }1$. Therefore $w\cdot \lambda -\lambda$ cannot have all nonnegative integer simple-root coordinates; if that coordinate is nonreal it is not even in the real root lattice, and if real it is negative. No distinct dot conjugate lies above $\lambda$. By [F3] this makes $\lambda$ maximal in $C_{\lambda }$, without claiming it is a greatest weight.

The auxiliary tensor. The simple pairings $⟨N\rho ,{\alpha }_i^{\vee }⟩=N$ make $N\rho$ dominant integral, so $E=L(N\rho )$ is finite-dimensional and $\mathfrak{h}$-semisimple by [F2]. Since $w_0$ reverses the positive roots, $w_0\rho =-\rho$, and the dual-highest-weight formula gives $E^{\ast }\cong L(-w_{0}N\rho )=E$. This argument concerns $N\rho$ only and imposes no integrality on $\mu$ or $\lambda$.

Projectivity in the actual block. The Verma $M(\lambda )$ is indecomposable: in a direct-sum decomposition, its one-dimensional highest weight space lies in exactly one summand, and its highest vector generates the whole Verma, so all other summands vanish. The block decomposition in [F4] consequently places $M(\lambda )$ entirely in the block containing its simple quotient $L(\lambda )$, namely $C_{\lambda }$. This class is a finite ideal of itself by [F3], and step 1.1 makes $\lambda$ maximal there. Apply [F4] to obtain projectivity in ${\mathcal{O}}_{C_{\lambda }}$; the exact block-projection adjunction of [F4] makes it projective in $\mathcal{O}$. Step 2.1 and [F5] then make $E\otimes M(\lambda )$ projective in $\mathcal{O}$.

The product of a highest-weight vector $e\in E^{\ast }=E$ of weight $N\rho$ and a highest-weight vector $v\in L(\mu )$ of weight $\mu$ is nonzero of weight $N\rho +\mu =\lambda$ and is annihilated by ${\mathfrak{n}}^{+}$; by [F6] it is the image of a nonzero element of ${\mathrm{Hom}}_{\mathcal{O}}(E\otimes M(\lambda ),L(\mu ))$, so that Hom-space is nonzero.

By [F7] there is a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}({\mathrm{pr}}_C(E\otimes M(\lambda )),L(\mu ))\cong {\mathrm{Hom}}_{\mathcal{O}}(E\otimes M(\lambda ),L(\mu ))\mathrm{\ne }0$, since $L(\mu )\in {\mathcal{O}}_C$. The object ${\mathrm{pr}}_C(E\otimes M(\lambda ))$ lies in ${\mathcal{O}}_C$ and is projective in $\mathcal{O}$ by step 3.1 and the first part of [F4]; a nonzero morphism from it to the simple object $L(\mu )$ is an epimorphism, so ${\mathcal{O}}_C$ contains a projective object mapping onto $L(\mu )$, as claimed.