I'm currently stuck in the proof of Proposition A.2.3. on page 267 of Monomial Ideals by Herzog and Hibi.
First we consider the following commutative diagram $\require{AMScd}$ \begin{CD} F @>{\varepsilon}>> U\\ @V \alpha VV @VV \varphi V\\ G @>{\eta}>> V \end{CD} Here $U,V,F,G$ are $S$-modules, $S=K[x_1,\dots,x_n]$ is the polynomial ring with coefficients over a field $K$. $F$ and $G$ are free f.g. modules, $\varepsilon$ and $\eta$ are homogeneous surjective homomorphism with $\ker(\varepsilon)\subseteq\mathfrak{m}F$ and $\ker(\eta)\subseteq\mathfrak{m}G$, and $\mathfrak{m}=(x_1,\dots,x_n)$ is the unique maximal omogeneous ideal of $S$.
On page 268, we consider the following commutative diagram $\require{AMScd}$ \begin{CD} F/\mathfrak{m}F @>{\overline{\varepsilon}}>> U/\mathfrak{m}U\\ @V \overline{\alpha} VV @VV \overline{\varphi} V\\ G/\mathfrak{m}G @>{\overline{\eta}}>> V/\mathfrak{m}V \end{CD} where the maps are the obvious one. By the costruction $\overline{\varepsilon},\overline{\varphi},\overline{\eta}$ are isomorphism, thus $\overline{\alpha}$ is an isomorphism. In order to complete the proof, one must show that $\alpha$ is also an isomorphism. The book only mentions that a homogeneous version of the Nakayama Lemma must be used. This is where I'm stuck.
This is the version I thought to use:
If $M$ is finitely generated graded $S$-module and $m_1,\dots, m_t\in M$ generate $M/\mathfrak{m}M$, then $m_1,\dots,m_t$ generate $M$.
I also know by basic tensor product isomorphism that $M/\mathfrak{m}M\cong K\otimes_S M$ for any $S$-module $M$. I don't understand how to use this to prove that $\ker(\alpha)=0$ for example.
$\newcommand{\m}{\mathfrak m}$ Here's a useful corollary of homogeneous Nakayama:
As $\bar\alpha$ is an isomorphism, it takes a basis of $F/\m F$ to a basis of $G/\m G$. Since $F$ and $G$ are finite free, the above corollary implies that $\alpha$ maps a basis of $F$ to a basis of $G$.