I am trying to understand the proof of algebraic independence of the characteristic classes of surface bundles, as outlined in Morita's book "Geometry of characteristic classes" (Theorem 4.19, page 163) and the original article "characteristic classes of surface bundles". I understand the construction of gluing bundles and the homorphism $\mathcal{M}_{g_1,1}\times\dots\times\mathcal{M}_{g_k,1}\rightarrow\mathcal{M}_{g,1}$, but I am stuck at the end, when it is said that it suffices to apply Künneth theorem to conclude. I do not understand how to do so, and how it yields algebraic independence. I also do not see where the assumption that $g\geq 3n$ is used. Could someone explain this last part? Or maybe give references for a proof with more details?
Thank you for your help!
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I think I found the answer. There was nothing deep apparently, it was just about looking carefully at the Künneth decomposition.
I am only going to prove a weaker version of it, with no precise lower bound for $g$. I did not see where the precise lower bound $g\geq 3n$ comes from.
Fix $n\geq 1$. For each $i=1,\dots,n$, let $M_i$ be an orientable closed $2i$-dimensional manifold and $E_i\xrightarrow{\pi_i}M_i$ a $\Sigma_{g_i}$-bundle with $0\neq e_i(\pi_i)\in H^{2i}(M_i;\mathcal{Q})$.
For each $i=1,\dots,n$, let $d_i\geq 1$ be the smallest integer such that $id_i\geq n$. Then let $M:=\prod_{i=1}^n M_i^{d_i}$ and $E\xrightarrow{\pi} M$ be the gluing of $d_1$ copies of $\pi_1$, $d_2$ copies of $\pi_2$, ..., $d_n$ copies of $\pi_n$ (see p. 170 in Morita's book). It is a $\Sigma_g$-bundle for $g=\sum_{i=1}^n id_i$. By (4.12) from Morita, we have \begin{equation} e_j(\pi)=\sum_{i=1}^{n}\sum_{k=1}^{d_i}p_{i,k}^*(e_j(\pi_i))\in H^{2j}(M;\mathcal{Q}) \end{equation} where $p_{i,k}\colon M\rightarrow M_i$ is the projection onto the $k$-th copy of $M_i$. We can make the sum start at $i=j$ since $H^{2j}(M_i;\mathcal{Q})=0$ for $i<j$, for dimension reason.
Let $p(e_1,\dots,e_n)\in\mathcal{Q}[e_1,\dots,e_n]$ be any nontrivial polynomial of degree $\leq 2n$. We want to show that its image in $H^*(\mathcal{M}_{g,1};\mathcal{Q})$ is nonzero. To do this, we show that, for $E\xrightarrow{\pi} M$, the polynomial $p(e_1,\dots,e_n)(\pi)\in H^*(M;\mathcal{Q})$ is nonzero. It suffices to consider each homogeneous part separately, hence we can assume that $p(e_1,\dots,e_n)$ is homogeneous of degree $2n$.
Let $i\in\{1,\dots,n\}$ be minimal such that $e_i$ appears in $p(e_1,\dots,e_n)$. It appears in a monomial of the form $e_i^{f_i}\cdot\dots\cdot e_n^{f_n}$ with $f_i\geq 1$, $f_{i+1},\dots,f_n\geq 0$ and $if_i+\dots+nf_n=n$ (note that, if $i<n$, then $f_n=0$). For each $j=i,\dots,n$, we have $2j f_j\leq 2n$, hence $f_j\leq d_j$ by the choice of $d_j$.
Under the Künneth isomorphism, $H^{2n}(M;\mathcal{Q})$ is a direct sum of several tensor products of cohomology groups of the $M_i$'s. The projection of $p(e_1,\dots,e_n)(\pi)$ to the direct summand $$H^{2i}(M_i;\mathcal{Q})^{\otimes f_i}\otimes\dots\otimes H^{2n}(M_n;\mathcal{Q})^{\otimes f_n}$$ is then given by $$e_i(\pi_i)^{\otimes f_i}\otimes\dots\otimes e_n(\pi_n)^{\otimes f_n},$$ which is nonzero by the choice of the $\pi_j$'s. Therefore $p(e_1,\dots,e_n)$ is nonzero in $H^{2n}(\mathcal{M}_{g,1};\mathcal{Q})$. This concludes the proof.