Invariants of graded space is the same as the graded space of invariants

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I am stuck on Theorem $3.5.1$ of the following book. The author says "In particular, we see that the map $(3.5.1)$ is compatible with the natural filtrations, and the corresponding map of associated graded spaces factors as follows $$ \operatorname{gr}\mathbb{C}[S_n]_{n \leq -2}|0\rangle \simeq \mathbb{C}[\bar{P}_n]_{n\leq -2} \to (\operatorname{gr}V_{\kappa_c}(\mathfrak{sl}_2))^{\mathfrak{sl}_2[[t]]} \simeq \operatorname{Inv}(\mathfrak{sl}_2)^*[[t]]$$ where the first map sends $S_n \mapsto \bar{P}_{1,n}$."

First of all I think that $\mathbb{C}[\bar{P}_n]_{n\leq -2}$ should be $\mathbb{C}[\bar{P}_{1, \, n}]_{n\leq -1} $, so that Theorem $(3.4.2)$ he refers to actually holds. Therefore the map should send $S_n \mapsto \bar{P}_{1,n+1}$, and this would also be consistent with what he notes some lines above: the symbol of the monomial $S_{n_1}\dots S_{n_m}|0\rangle$ is $\bar{P}_{1,n_1+1}\dots \bar{P}_{1, n_m+1}$. Am I correct or am I just missing something?

Then another question: the map $(3.5.1)$ he is referring to is the inclusion $\mathbb{C}[S_n]_{n \leq -2}|0\rangle \to V_{\kappa_c}(\mathfrak{sl}_2)^{\mathfrak{sl}_2[[t]]}$. Of course the inclusion $\mathbb{C}[S_n]_{n \leq -2}|0\rangle \to V_{\kappa_c}(\mathfrak{sl}_2)$ induces a map $$\operatorname{gr}\mathbb{C}[S_n]_{n \leq -2}|0\rangle \to \operatorname{gr}V_{\kappa_c}(\mathfrak{sl}_2)$$ on the graded spaces. From what it looks like this map actually sends elements to $(\operatorname{gr}V_{\kappa_c}(\mathfrak{sl}_2))^{\mathfrak{sl}_2[[t]]}$? Is this in some way obvious?

Or is the reason the following: the map induces $\operatorname{gr}\mathbb{C}[S_n]_{n \leq -2}|0\rangle \to \operatorname{gr}(V_{\kappa_c}(\mathfrak{sl}_2)^{\mathfrak{sl}_2[[t]]})$ and the we have the equality $\operatorname{gr}(V_{\kappa_c}(\mathfrak{sl}_2)^{\mathfrak{sl}_2[[t]]}) = (\operatorname{gr}V_{\kappa_c}(\mathfrak{sl}_2))^{\mathfrak{sl}_2[[t]]}$?

I can't get my head around all of this, I hope someone will be able to help me.