Showing minimal graded free resolutions are isomorphic I'm currently reading Rogalski's notes on noncommutative projective algebraic geometry (which can be found here) and I'm currently trying to fill out the details of Lemma 1.24 (2). The step which I don't understand is why the map $h$ is an isomorphism. I understand that there is an isomorphism $P/PA_{\ge 1} \rightarrow Q/QA_{\ge 1}$, and how we can lift this to a map $h: P \rightarrow Q$, but I can't see either injectivity or surjectivity of $h$. I feel like graded Nakayama should be relevant, but can't see why.

Let $\varphi: P/PA_{\ge 1} \rightarrow Q/QA_{\ge 1}$ be the graded isomorphism and let $h:P \to Q$ resp. $g:Q \to P$ be graded lifts of $\varphi$ resp. $\varphi^{-1}$. Then $gh\equiv 
\operatorname{id}_P \operatorname{mod} PA_{\ge 1}$ and $hg\equiv \operatorname{id}_Q \operatorname{mod} QA_{\ge 1}$. Hence it suffices to show that $gh, hg$ are isomorphisms. This follows from: 

Let $A$ be a positively graded graded ring with unit. Let $f: M \to M$ be an endomorphism of a positively graded $A$-module that induces the identity on $M/MA_{\ge 1}$. Then $f$ is an isomorphism and in degree $n$ the inverse is given by 
  $$f^{-1} = \sum_{k=0}^n(-1)^k\binom{n+1}{k+1}f^k.$$

Proof: Denote the right hand side by $R$. Hence 
$$\operatorname{id}-f\circ R = (\operatorname{id}-f)^{n+1}$$
Let $x \in M$ be of degree $n$. Since $f$ induces the identity on $M/MA_{\ge 1}$ we have $x-f(x) \in MA_{\ge 1}$, i.e. there are homogeneous $a_i^1\in A_{\ge 1}, x_i^1\in M$ with $\deg(x_i^1a_i^1)=n\,$ (in particular $\deg x_i^1 \le n-1)$ such that $x-f(x)=\sum_i x_i^1a_i^1$. Hence 
$$(\operatorname{id}-f\circ R)(x)=\sum_i (\operatorname{id}-f)^n(x_i^1)a_i^1.$$
By applying the same argument to $x_i^1$, we can write 
$$(\operatorname{id}-f\circ R)(x)=\sum_i (\operatorname{id}-f)^{n-1}(x_i^2)a_i^2.$$
with homogeneous $a_i^2,x_i^2$ and $\deg(x_i^2)\le n-2$. Continuing this process, we have in the last step
$$(\operatorname{id}-f\circ R)(x)=\sum_i (\operatorname{id}-f)(x_i^n)a_i^n$$
with $\deg(x_i^n) \le 0$. Once more write $(\operatorname{id}-f)(x_i^n)= \sum_j x_j^{n+1}a_j^{n+1}$ with homogeneous $a_j^{n+1}\in A_{\ge 1},x_j^{n+1}\in M$ and $\deg(x_j^{n+1}a_j^{n+1})=0$. But because $\deg(a_j^{n+1})\ge 1$, $x_j^{n+1}$ has negative degree, meaning $x_j^{n+1}=0$ since $M_{<0}=0$. Hence $(\operatorname{id}-f\circ R)(x)=0$. So in degree $n$: 
$$f \circ R = \operatorname{id}=R \circ f$$
where the last identity uses that $f$ commutes with $R$. qed
