As $U\left({𝔤}_{<0}\right)\text{-modules,}$ $M\left(\lambda \right)\cong U\left({𝔤}_{<0}\right)\text{.}$ Therefore, Proposition 2.0.5 implies that $M\left(\lambda \right)=\underset{\mu \le \lambda }{⨁}M{\left(\lambda \right)}^{\mu }$ and $M{\left(\lambda \right)}^{\mu }$ has a basis $\left\{x_{\left(t\right)}{v}^{+}\hspace{0.17em}\right|\hspace{0.17em}\left(t\right):t_1\ge \cdots \ge t_j>0,\lambda -\mu ={\beta }_{t_1}+\cdots +{\beta }_{t_j}\}\text{.}$

Since $𝔤$ has a regular triangular decomposition, $\text{dim}\hspace{0.17em}M{\left(\lambda \right)}^{\mu }<\infty \text{.}$ Therefore, $M\left(\lambda \right)\in 𝒪\text{.}$

$\square$

Since every element of $M$ is a finite sum of weight vectors, $M$ is generated by a finite sum of weight vectors. Let $\{m_1,\dots ,m_n\}$ be a minimal generating set of weight vectors for $M\text{.}$ Assume $m_i\in M^{{\gamma }_i}\text{.}$ Consider the set of submodules $$𝒦=\{K\subseteq M\hspace{0.17em}|\hspace{0.17em}{m}_n\notin K\}\text{.}$$ This set is nonempty (since $\left\{0\right\}\in 𝒦\text{)}$ and partially ordered by inclusion. We claim that this set has a maximal element. To prove this claim, let $K_1\subset K_2\subset \cdots$ be an increasing chain of submodules in $𝒦,$ and define $K=\bigcup _i{K}_i\text{.}$ Then $K$ is a submodule of $M\text{.}$

Since $\text{dim}\hspace{0.17em}{M}^{{\gamma }_n}<\infty ,$ there is some $k\in {\mathbb{Z}}_{>0}$ so that $K_i^{{\gamma }_n}=K_{i+1}^{{\gamma }_n}$ for all $i\ge k\text{.}$ Since $m_n\notin K_i$ for all $i\in {\mathbb{Z}}_{>0},$ this implies $m_n\notin K\text{.}$ Therefore, $K\in 𝒦,$ and so every increasing chain in $𝒦$ has a maximal element. Zorn's Lemma implies that $𝒦$ has a maximal element $\bar{K}\text{.}$ By construction, any submodule $M\prime$ of $M$ which properly contains $\bar{K}$ is not in $𝒦$ and so must contain $m_n\text{.}$ Because we have chosen the generating set $\{m_1,\dots ,m_n\}$ to be minimal, the submodule generated by $m_i$ $\text{(}i\ne n\text{)}$ is contained in $𝒦,$ and so $m_i\in \bar{K}\subset M\prime$ for $1\le i\le n-1\text{.}$ This implies $M\prime =M\text{.}$ Therefore, if $\bar{K}$ is a maximal proper submodule.

Any maximal proper submodule of $M$ must not contain at least one of the generators of $M\text{.}$ Since $M\left(\lambda \right)$ is generated by one element, this implies $M\left(\lambda \right)$ has a unique maximal proper submodule.

$\square$

Let $L\in 𝒪$ be a simple module. Since $L\in 𝒪,$ we can choose $\lambda \in {𝔥}^{*}$ so that $L^{\lambda }\ne 0$ and $L^{\mu }=0$ for all $\mu >\lambda \text{.}$ Let $0\ne v\in L^{\lambda }\text{.}$ Then $v$ is a highest weight vector, and so there is a map $\varphi :M\left(\lambda \right)\to L$ given by $\varphi :v^{+}\mapsto v\text{.}$ Since $L$ is irreducible, $\text{ker}\left(\varphi \right)=J\left(\lambda \right)\text{.}$

To show that the simple modules $L\left(\lambda \right)$ are distinct, suppose $\lambda ,\mu \in {𝔥}^{*}$ with $\lambda \ne \mu \text{.}$ We may assume $\lambda \nleqq \mu \text{.}$ Therefore, $L{\left(\mu \right)}^{\lambda }=0$ and so $L\left(\mu \right)\ne L\left(\lambda \right)\text{.}$

$\square$

Let ${\left\{v_i\right\}}_{i=1}^n$ be a generating set for $P\text{.}$ Since $P\in 𝒪,$ every element of $P$ can be written as a finite sum of weight vectors. Therefore, we assume $v_i\in P^{{\nu }_i}$ for some ${\nu }_i\in {𝔥}^{*}\text{.}$

Suppose $P$ can be written as a finite direct sum of nonzero submodules, $P=P_1\oplus \cdots \oplus P_m\text{.}$ Write $v_i=v_i^1+\cdots +v_i^m$ where $v_i^j\in P_j\text{.}$ Then ${\left\{v_i^j\right\}}_{i=1}^n$ generates $P_j\text{.}$ This implies that for each $j$ there is at least one $1\le i\le n$ such that $v_i^j\ne 0\text{.}$ Therefore, $$m\le \sum _{i,j}\text{dim}\hspace{0.17em}{P}_j^{{\nu }_i}=\sum _i\text{dim}\hspace{0.17em}{P}^{{\nu }_i}<\infty$$ since $P\in 𝒪\text{.}$ Therefore, we can continue to decompose $P$ until the summands $P_i$ are indecomposable; since $m$ is bounded, the number of terms in this decomposition will be finite.

$\square$

We prove this claim by inducting on $\sum _{\mu \ge \lambda }\text{dim}\hspace{0.17em}{M}^{\mu }\text{.}$ (This sum is always finite since $\text{dim}\hspace{0.17em}{M}^{\mu }<\infty$ and there are a finite number of weights between any two weights.)

If $\sum _{\mu \ge \lambda }\text{dim}\hspace{0.17em}{M}^{\mu }=0,$ then $M\supseteq 0$ is a local composition series at $\lambda \text{.}$ Suppose $\sum _{\mu \ge \lambda }\text{dim}\hspace{0.17em}{M}^{\mu }=n,$ and assume the result holds for all modules $M\prime \in 𝒪$ with $\sum _{\mu \ge \lambda }\text{dim}{\left(M\prime \right)}^{\mu }<n\text{.}$ Choose $\mu \in {𝔥}^{*}$ maximal so that $M^{\mu }\ne 0$ and $\mu \ge \lambda \text{.}$ Let $0\ne v\in M^{\mu }\text{.}$ By the maximality of $\mu ,$ $v$ must be a highest weight vector. Let $V=U\left(𝔤\right)v$ and let $U$ be the maximal submodule of $V\text{.}$ Then $V/U\cong L\left(\mu \right)\text{.}$

We have $\sum _{\mu \ge \lambda }\text{dim}{(M/V)}^{\mu }<n$ and $\sum _{\mu \ge \lambda }\text{dim}\hspace{0.17em}{U}^{\mu }<\sum _{\mu \ge \lambda }\text{dim}\hspace{0.17em}{V}^{\mu }\le n\text{.}$ Therefore, by the inductive hypothesis, both $M/V$ and $U$ have local composition series at $\lambda \text{:}$ $$M/V=V_k\supseteq \cdots \supseteq V_1\supseteq V_0=0$$ $$U=U_l\supseteq \cdots \supseteq U_1\supseteq U_0=0\text{.}$$ For each $0\le i\le k,$ let $V\subseteq M_i\subseteq M$ be a submodule of $M$ so that $M_i/V\cong V_i\text{.}$ Then, $$M=M_k\supseteq \cdots \supseteq M_0=V\supseteq U=U_l\supseteq \cdots \supseteq U_0=0$$ is a local composition series for $M$ at $\lambda \text{.}$

$\square$

Let $$\begin{array}{rcl}{\displaystyle M}& =& {\displaystyle M_n\supseteq M_{n-1}\supseteq \cdots \supseteq M_0=0}\\ {\displaystyle M}& =& {\displaystyle M_m^{\prime }\supseteq M_{m-1}^{\prime }\supseteq \cdots \supseteq M_0^{\prime }=0}\end{array}$$ be Verma composition series for $M$ with $M_i/M_{i-1}\cong M\left({\lambda }_i\right),$ ${\lambda }_i\in {𝔥}^{*}$ and $M_i^{\prime }/M_{i-1}^{\prime }\cong M\left({\mu }_j\right),$ ${\mu }_j\in {𝔥}^{*}\text{.}$ Then, $$\begin{array}{cc}{\displaystyle \sum _{i=1}^n\text{ch}\hspace{0.17em}M\left({\lambda }_i\right)=\text{ch}\hspace{0.17em}M=\sum _{j=1}^m\text{ch}\hspace{0.17em}M\left({\mu }_j\right)\text{.}}& \text{(2.2)}\end{array}$$ Choose ${\lambda }_k$ maximal in $\{{\lambda }_1,\dots ,{\lambda }_m\}\text{.}$ This implies the coefficient of $e^{{\lambda }_k}$ in (2.2) is $1$ and so ${\mu }_l={\lambda }_i$ for some $l\text{.}$ Therefore, $\sum _{i\ne k}\text{ch}\hspace{0.17em}M\left({\lambda }_i\right)=\sum _{j\ne l}\text{ch}\hspace{0.17em}M\left({\mu }_j\right)\text{.}$ Induct on $m$ to get the result.

$\square$

Let $M\in 𝒪\text{.}$ Choose ${\lambda }_1,\dots ,{\lambda }_n\in {𝔥}^{*}$ so that $M^{\mu }\ne 0$ only for $\mu \le {\lambda }_i\text{.}$ We may assume that any weight $\mu$ of $M$ is comparable to only one ${\lambda }_i\text{.}$ (If $\mu \le {\lambda }_i$ and $\mu \le {\lambda }_j,$ then we can replace ${\lambda }_i$ and ${\lambda }_j$ by some $\lambda \in {𝔥}^{*}$ so that $\lambda >{\lambda }_i,{\lambda }_j\text{.)}$ Let $\mu$ be a weight of $M$ and suppose $\mu \le {\lambda }_i\text{.}$ Then define $d\left(\mu \right)=\text{ht}({\lambda }_i-\mu )\text{.}$

Choose ${\mu }_1\in {𝔥}^{*}$ a weight of $M$ so that $d\left({\mu }_1\right)$ is minimal. This implies that any nonzero vector in $M^{{\mu }_1}$ is a highest weight vector. Therefore, choose $0\ne v_1\in M^{{\mu }_1}$ and let $M^1$ be the $𝔤\text{-submodule}$ of $M$ generated by $v_1\text{.}$

Now choose a weight ${\mu }_2$ of $M/M^1$ so that $d\left({\mu }_2\right)$ is minimal. Then we can choose $0\ne v_2\in {(M/M^1)}^{{\mu }_2}\text{.}$ Let $M^2\supseteq M^1$ be the submodule of $M$ so that $M^2/M^1$ is generated by $v_2\text{.}$ In this way we inductively define $M^i\text{.}$ The following picture illustrates this process of choosing the weights ${\mu }_i\text{:}$ $$\begin{array}{c} λi λj d(μ1) d(μ2) d(μ3) μ1 μ2 μ3 \end{array}$$ In this way, we construct a filtration $0=M^0\subseteq M^1\subseteq M^2\subseteq \cdots$ such that each quotient is a highest weight module. Since $Q$ is finitely generated, there are only a finite number of weights $\mu$ of $M$ such that $d\left(\mu \right)$ has a given value. This, along with the fact that $\text{dim}\hspace{0.17em}{M}^{\mu }<\infty ,$ show that the construction outlined above yields a filtration so that $M=\bigcup _{i=1}^{\infty }{M}^i\text{.}$

$\square$

Let ${\left\{w_i\right\}}_{i=1}^n$ be a basis of weight vectors for $W\left(\mu \right)\text{.}$ (This set is finite since $\text{dim}\hspace{0.17em}W{\left(\mu \right)}^{\nu }\le \text{dim}\hspace{0.17em}U{\left({𝔤}_{>0}\right)}^{\nu }<\infty$ for all $\nu$ and $W{\left(\mu \right)}^{\nu }=0$ unless $\mu \le \nu \le \lambda \text{.)}$ As a $U\left({𝔤}_{<0}\right)\text{-module,}$ $\stackrel{\sim }{P}\left(\mu \right)$ is (freely) generated by $1\otimes w_i\text{.}$ Therefore, $\stackrel{\sim }{P}\left(\mu \right)$ is finitely generated as a $𝔤\text{-module}$ This implies $\stackrel{\sim }{P}{\left(\mu \right)}^{\nu }\ne 0$ only if $\nu \le \lambda$ and $\stackrel{\sim }{P}{\left(\mu \right)}^{\nu }<\infty \text{.}$

$\square$

Let $M$ and $N$ be modules in $𝒪\left(\lambda \right)$ such that $$\begin{array}{ccc}& & \stackrel{\sim }{P}\left(\mu \right)\\ & & \downarrow \varphi \\ N& \stackrel{\psi }{⟶}& M& ⟶& 0\text{.}\end{array}$$ Note that elements of $\stackrel{\sim }{P}\left(\mu \right)$ are sums of elements of the form $x\otimes \bar{y},$ where $x\in U\left(𝔤\right)$ and $\bar{y}\in W\left(\mu \right)$ is the image of $y\in U\left({𝔤}_{>0}\right)\text{.}$ Suppose $\varphi (1\otimes \bar{1})=m_0\in M\text{.}$ Let $n_0\in N$ be such that $\psi \left(n_0\right)=m_0\text{.}$ Define a map $\varphi \prime :\stackrel{\sim }{P}\left(\mu \right)\to N$ by $$\varphi \prime (x\otimes \bar{y})=xy{n}_0\text{.}$$ If this map is well-defined, it is clearly a $𝔤\text{-module}$ homomorphism such that $\psi \otimes \varphi \prime =\varphi \text{.}$ Therefore we check that $\varphi \prime$ is well-defined. Suppose that $\bar{y}=\bar{y}\prime \text{.}$ Then $y=y\prime +u$ for some $u\in \underset{\nu \nleqq \lambda }{⨁}U{\left({𝔤}_{>0}\right)}^{\nu }\text{.}$ This implies $$xy{n}_0=xy\prime n_0+xu{n}_0\text{.}$$ However, $n_0\in N^{\mu }$ and so $u{n}_0\in \underset{\nu \nleqq \lambda }{⨁}{N}^{\nu }=0\text{.}$

$\square$

Note that $\stackrel{\sim }{P}\left(\mu \right)$ is a free $U\left({𝔤}_{<0}\right)\text{-module.}$

Let ${\left\{w_i\right\}}_{i=1}^n$ be a basis of weight vectors for $W\left(\mu \right)\text{.}$ Assume that $w_i\in W{\left(\mu \right)}^{{\gamma }_i}$ and ${\gamma }_i>{\gamma }_j$ implies $i>j\text{.}$ Let $M_i$ be the submodule of $\stackrel{\sim }{P}\left(\mu \right)$ generated by $1\otimes w_i\text{.}$ Then the following is a Verma composition series for $\stackrel{\sim }{P}\left(\mu \right)\text{:}$ $$0\subseteq M_n\subseteq M_{n-1}+M_n\subseteq \cdots \subseteq M_1+\cdots +M_n=\stackrel{\sim }{P}\left(\mu \right)\text{.}$$ Also, for $\mu \le \nu \le \lambda ,$ $$[\stackrel{\sim }{P}\left(\mu \right):M\left(\nu \right)]=\text{dim}\hspace{0.17em}W{\left(\mu \right)}^{\nu }=\text{dim}\hspace{0.17em}M{\left(\nu \right)}^{\mu }\text{.}$$ Since $\stackrel{\sim }{P}\left(\mu \right)$ is projective, $$\text{dim}\hspace{0.17em}M{\left(\nu \right)}^{\mu }=\text{dim}\hspace{0.17em}{\text{Hom}}_{𝔤}(\stackrel{\sim }{P}\left(\mu \right),M\left(\nu \right))\text{.}$$

$\square$

Let $\mu \le \lambda \text{.}$ From Lemma 2.1.5, we have $\stackrel{\sim }{P}\left(\mu \right)=\underset{1\le i\le n}{⨁}{P}_i,$ where each $P_i$ is an indecomposable projective module. Moreover, each $P_i$ is generated by one element, namely the projection of $1\otimes \bar{1}\in \stackrel{\sim }{P}\left(\mu \right)\text{.}$ Lemma 2.1.3 shows that $P_i$ has a maximal proper submodule. Since $P_i$ is generated by one element, this submodule is unique.

Now consider any finitely generated indecomposable projective module $P\text{.}$ Let $\{v_1,\dots ,v_n\}$ be a generating set for $P$ so that $v_i\in P^{{\nu }_i}\text{.}$ Then there is a surjective map $\varphi :\sum _i\stackrel{\sim }{P}\left({\nu }_i\right)\to P$ defined by $\varphi \left(v_{{\nu }_i}\right)=v_i,$ where $v_{{\nu }_i}=1\otimes \bar{1}\in \stackrel{\sim }{P}\left({\nu }_i\right)\text{.}$ Since $P$ is projective and indecomposable, it must be isomorphic to an indecomposable summand of $\sum _i\stackrel{\sim }{P}\left({\nu }_i\right)\text{.}$ The previous paragraph implies that $P$ has a unique maximal submodule.

$\square$

Let $\mu \le \lambda \text{.}$ Then $L\left(\mu \right)$ is a quotient of $\stackrel{\sim }{P}\left(\mu \right)$ and therefore must be a quotient of a projective indecomposable summand $P$ of $\stackrel{\sim }{P}\left(\mu \right)\text{.}$ Then $P$ is a projective cover of $L\left(\mu \right),$ which we denote by $P\left(\mu \right)\text{.}$

The previous proposition shows that each finitely generated indecomposable module $P$ has a unique irreducible quotient, and so $P\cong P\left(\mu \right)$ for some $\mu \le \lambda \text{.}$

$\square$

We begin by showing $[M\left(\mu \right):L\left(\nu \right)]=\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M\left(\mu \right))\right)\text{.}$ According to Lemma 2.2.1, $M\left(\mu \right)$ has a local composition series at $\nu ,$ $M\left(\mu \right)=M_1\supseteq \cdots \supseteq M_n=0\text{.}$ Lemma 1.3.3 implies $$\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M\left(\nu \right))\right)=\sum _{k=1}^{n-1}\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M_k/M_{k+1})\right)\text{.}$$ Consider $\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M_k/M_{k+1})\right)\text{.}$ By definition of a local composition series $M_k/M_{k+1}\cong L\left(\gamma \right)$ for some $\gamma \ge \nu$ or ${(M_k/M_{k+1})}^{\gamma }=0$ for all $\gamma \ge \nu \text{.}$ Now, $$\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),L\left(\gamma \right))\right)=\{\begin{array}{ll}0& \text{if}\hspace{0.17em}\nu \ne \gamma \\ ℂ& \text{if}\hspace{0.17em}\nu =\gamma \end{array}$$ since $P\left(\nu \right)$ has a unique maximal submodule $L\left(\nu \right)\text{.}$ Suppose ${(M_k/M_{k+1})}^{\gamma }=0$ for all $\gamma \ge \nu \text{.}$ Since $L\left(\nu \right)$ is the unique irreducible quotient of $P\left(\nu \right)$ and $L{\left(\nu \right)}^{\nu }\ne 0,$ any homomorphism $\varphi :P\left(\nu \right)\to M_k/M_{k+1}$ must be trivial. Therefore, $\text{dim}\hspace{0.17em}{\text{Hom}}_{𝔤}(P\left(\nu \right),M_k,M_{k+1})=0\text{.}$

We now have $$\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M\left(\mu \right))\right)=\sum _{k=1}^n\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M_k/M_{k+1})\right)=[M\left(\mu \right):L\left(\nu \right)]\text{.}$$

Next we show $\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right),M\left(\mu \right))\right)=[P\left(\nu \right),M\left(\mu \right)]\text{.}$ Recall $\text{dim}\left({\text{Hom}}_{𝔤}(\stackrel{\sim }{P}\left(\nu \right),M\left(\mu \right))\right)=[\stackrel{\sim }{P}\left(\nu \right),M\left(\mu \right)]$ and $\stackrel{\sim }{P}\left(\nu \right)=\underset{\gamma }{⨁}{m}_{\nu }\left(\gamma \right)P\left(\gamma \right)$ for some $m_{\nu }\left(\gamma \right)\in {\mathbb{Z}}_{\ge 0}\text{.}$ Therefore, $$\begin{array}{cc}{\displaystyle \begin{array}{rcl}{\displaystyle \underset{\gamma }{⨁}{m}_{\nu }\left(\gamma \right)\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\gamma \right),M\left(\mu \right))\right)}& =& {\displaystyle \text{dim}\left({\text{Hom}}_{𝔤}(\stackrel{\sim }{P}\left(\nu \right),M\left(\mu \right))\right)}\\ & =& {\displaystyle [\stackrel{\sim }{P}\left(\nu \right):M\left(\mu \right)]}\\ & =& {\displaystyle \underset{\gamma }{⨁}{m}_{\nu }\left(\gamma \right)[P\left(\gamma \right):M\left(\mu \right)]}\end{array}}& \text{(2.3)}\end{array}$$ Observe that $$\text{dim}\left({\text{Hom}}_{𝔤}(\stackrel{\sim }{P}\left(\nu \right),L\left(\omega \right))\right)=\underset{\gamma }{⨁}{m}_{\nu }\left(\gamma \right)\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\gamma \right),L\left(\omega \right))\right)=m_{\nu }\left(\omega \right)\text{.}$$ On the other hand, $\text{dim}\left({\text{Hom}}_{𝔤}(\stackrel{\sim }{P}\left(\nu \right),L\left(\omega \right))\right)=\text{dim}\hspace{0.17em}L{\left(\omega \right)}^{\nu }$ since $\stackrel{\sim }{P}\left(\nu \right)$ acts like a free module in $𝒪\left(\lambda \right)\text{.}$ Therefore, $m_{\nu }\left(\nu \right)=1$ and $m_{\nu }\left(\gamma \right)=0$ unless $\nu \le \gamma \text{.}$ Using Equation 2.3, we can then induct on $\text{ht}(\lambda -\nu )$ to show $[P\left(\nu \right):M\left(\mu \right)]=\text{dim}\left({\text{Hom}}_{𝔤}(P\left(\nu \right):M\left(\mu \right))\right)\text{.}$

$\square$

We first claim $\lambda \ngtr \mu$ implies ${\text{Ext}}^1(M\left(\mu \right),L\left(\lambda \right))=0\text{.}$ To prove this, assume $\lambda \ngtr \mu$ and let $$0⟶L\left(\lambda \right)⟶N\stackrel{\varphi }{⟶}M\left(\mu \right)⟶0$$ be an extension of $M\left(\mu \right)$ by $L\left(\lambda \right)\text{.}$ Let $v\in N^{\mu }$ so that $\varphi \left(v\right)=v^{+}\in M\left(\mu \right)\text{.}$ If $\lambda \ngtr \mu ,$ then $N^{\gamma }=0$ for all $\gamma >\mu$ and so $v$ is a highest weight vector in $N\text{.}$ Therefore, we have a homomorphism $\psi :M\left(\mu \right)\to N$ given by $\psi \left(v^{+}\right)=v\text{.}$ Then $\psi \circ \varphi ={\text{id}}_{M\left(\lambda \right)}$ and so the sequence splits. This implies ${\text{Ext}}^1(M\left(\mu \right),L\left(\lambda \right))=0\text{.}$

We now assume ${\text{Ext}}^1(M\left(\mu \right),L\left(\lambda \right))\ne 0\text{.}$ Since $\lambda >\mu ,$ $M\left(\mu \right)\in 𝒪\left(\lambda \right)\text{.}$ Using Theorem 2.3.6. we only need to show $[P\left(\mu \right),M\left(\lambda \right)]>0\text{.}$

Since $\stackrel{\sim }{P}\left(\mu \right)$ has a Verma composition series (Proposition 2.3.3) and $P\left(\mu \right)$ is a direct summand of $\stackrel{\sim }{P}\left(\mu \right),$ $P\left(\mu \right)$ has a Verma composition series $P\left(\mu \right)=P_0\supseteq \cdots \supseteq P_n=0$ such that $P\left(\mu \right)/P_1\cong M\left(\mu \right)\text{.}$ We then have the following diagram $$\begin{array}{ccccccc}& & & & & & P\left(\mu \right)\\ & & & & & & \downarrow \varphi \\ 0& ⟶& L\left(\lambda \right)& ⟶& N& \stackrel{\psi }{⟶}& M\left(\mu \right)& ⟶& 0\text{.}\end{array}$$ Since $P\left(\mu \right)$ is projective, there is a homomorphism $\varphi \prime :P\left(\mu \right)\to N$ such that $\psi \circ \varphi \prime =\varphi \text{.}$ Then, $\varphi \prime \left(P_1\right)\subseteq \text{Ker}\hspace{0.17em}\psi \cong L\left(\lambda \right)\text{.}$ We must have either $\varphi \prime \left(P_1\right)=0$ or $\varphi \prime \left(P_1\right)\cong L\left(\lambda \right)\text{.}$ If $\varphi \prime \left(P_1\right)=0,$ $\varphi \prime$ induces a map $\varphi \prime :M\left(\mu \right)\to N$ and the sequence splits. Therefore, $\varphi \prime \left(P_1\right)\cong L\left(\lambda \right)\text{.}$ Applying $\varphi \prime$ to the Verma composition series for $P\left(\mu \right),$ we have $$\varphi \prime \left(P_1\right)\cong L\left(\lambda \right)\supseteq \cdots \supseteq \varphi \prime \left(P_n\right)=0\text{.}$$ Therefore, there is some $1\le k\le n-1$ such that $\varphi \prime \left(P_k\right)\cong L\left(\lambda \right)$ and $\varphi \prime \left(P_{k+1}\right)=0\text{.}$ Then $P_k/P_{k+1}\cong M\left(\lambda \right)\text{.}$

$\square$

We have $\lambda >\mu \text{.}$ Let $N$ be the maximal submodule of $M\text{.}$ Then we have a short exact sequence $$0⟶N⟶M⟶L\left(\lambda \right)⟶0\text{.}$$ Proposition 1.3.4 implies that ${\text{Ext}}^1(M\left(\mu \right),L\left(\lambda \right))\ne 0$ or ${\text{Ext}}^1(M\left(\mu \right),N)>0\text{.}$ If ${\text{Ext}}^1(M\left(\mu \right),L\left(\lambda \right))\ne 0,$ the proposition holds. Therefore assume that ${\text{Ext}}^1(M\left(\mu \right),N)\ne 0\text{.}$ Let $$N=N_0\subseteq N_1\subseteq \cdots$$ be a local composition series for $N\text{.}$ Then, by Proposition 1.3.4, ${\text{Ext}}^1(M\left(\mu \right),N_i/N_{i+1})\ne 0$ for some $i\text{.}$

Recall that $N_i/N_{i+1}\cong L\left(\nu \right)$ for some $\lambda \ge \nu \ge \mu$ or ${(N_i/N_{i+1})}^{\nu }=0$ for all $\nu \ge \mu \text{.}$ Suppose ${(N_i/N_{i+1})}^{\nu }=0$ for all $\nu \ge \mu \text{.}$ We claim this implies ${\text{Ext}}^1(M\left(\mu \right),N_i/N_{i+1})=0,$ a contradiction. Let $$0⟶N_i/N_{i+1}⟶N\prime ⟶M\left(\mu \right)⟶0$$ be a short exact sequence. Choose $v\in N\prime$ so that $v\mapsto v^{+}\in M\left(\mu \right)\text{.}$ Then $v\in {\left(N\prime \right)}^{\mu }\text{.}$ Since ${(N_i/N_{i+1})}^{\nu }=0$ and $M{\left(\mu \right)}^{\nu }=0$ for $\nu >\mu ,$ ${\left(N\prime \right)}^{\nu }=0\text{.}$ This implies that $v$ is a highest weight vector. We then have a map $M\left(\mu \right)\to N$ given by $v^{+}\mapsto v\text{.}$ Thus, the sequence splits, and so ${\text{Ext}}^1(M\left(\mu \right),N_i/N_{i+1})=0\text{.}$