I am having problems to understand the first lines of the proof of Proposition 12.1 in "Rational Homotopy Theory" by Felix-Halperin-Thomas.
Suppose $(A,d)$ is a cdga over $k$ with $H^0(A)=k$, where $k$ field of characteristic nonzero. Then one wants to build a quasi-isomorphism $$m: (\Lambda V, d) \to (A,d)$$ from a Sullivan algebra to $A$.
The proof in the book starts as follows:
"We construct this so that $V$ is the direct sum of graded subspaces $V_k$ , $k \geq 0$, with $d=0$ in $V_0$ and $d: V_k \to \Lambda (V_0 > \oplus \cdots \oplus V_{k-1})$. Choose $m_0 : (\Lambda V_0, 0) \to > (A,d)$ so that $$H(m_0): V_0 \overset{\cong}{\to} H^+(A).$$ Since $H^0(A)=k$, $H(m_0)$ is surjective."
Could someone please elaborate on this? More precisely,
How to choose $V_0$?
What are exactly the two maps $H(m_0)$ that appear in the proof? Does $H(m_0): V_0 \overset{\cong}{\to} H^+(A)$ refer to $H^+(m_0):H^+(\Lambda V_0) =V_0 \overset{\cong}{\to} H^+(A)$? (I don't think so as $H^+(\Lambda V_0) =V_0$ does not hold, I think...).
What is the surjective $H(m_0)$? Is it $H(m_0): H(\Lambda V_0) \to H(A)$ ?
Perhaps it is just that there is a small typo in the proof (clearly $H(m_0)$ cannot be first an isomorphism to deduce later that it is surjective), but I am not able to see what the correct sentence should be.
Define $V_0$ as a graded vector space on a basis $\{v_i\}_{i\in I}$ in bijection with a linear basis $\{[a_i]\}_{i\in I}$ of $H^+(A)$.
$m_0$ is mapping each $v_i$ to a chosen representative $x_i\in [a_i]$. This produces a well-defined cdga map $\Lambda V_0 \to A$, which induces a map in cohomology $H(m_0)$ whose restriction to $V_0$ is a bijection by construction (that's the $H(m_0):V_0 \to H^+(A)$ part). Note $H^*(\Lambda V_0) \cong \Lambda V_0$ because $d=0$ on $V_0$.
The map $H(m_0)$ is surjective on $H^+(A)$ by construction. Since $H^0(A)\cong k$, the final resulting map is surjective because we can map $1\mapsto 1$.