{\mathbb{I}}\to {\mathbb{G}}$ be measurable and bounded. Let $\pi \colon {\mathbb{I}}^2 \to {\mathbb{I}}$ be the projection on the first coordinate, and consider the map $$A \circ \pi \colon ({\mathbb{I}}^2,m) \to {\mathbb{G}}.$$ Then $\Lambda_k (A \circ \pi) = \Lambda_k(A)$ for every $k \in {\mathbb{N}}$.

The idea of the proof is to approximate $\pi$ by something invertible and to use remark \[rem.isomorphism\] and the continuity property from lemma \[l.cont\].

It is clear that $\Lambda_k(A \circ \pi) \le \Lambda_k(A)$, because $\Lambda_k(A \circ \pi, T \times {\mathit{id}}) = \Lambda_k(A,T)$ for any $T \in {\mathit{Aut}}({\mathbb{I}},m)$. Fix $\delta>0$. Let $T \in {\mathit{Aut}}({\mathbb{I}}^2,m)$ be such that $\Lambda_k(A \circ \pi,T) < \Lambda_k(A\circ \pi) + \delta$. For $n\in{\mathbb{N}}$, define an isomorphism $P_n\colon  ({\mathbb{I}}^2,m) \to ({\mathbb{I}},m)$ such that if $I \subset {\mathbb{I}}$ is a dyadic interval with $|I|=2^{-n}$ then $P_n(I \times {\mathbb{I}}) = I$. Then the functions $\pi$ and $P_n\colon {\mathbb{I}}^2 \to {\mathbb{I}}$ are uniformly $2^{-n}$-close. This implies $L^1$-convergence: $$\lim_{n\to\infty} \| A \circ P_n - A \circ \pi \|_1 = 0.$$ Indeed, given ${\varepsilon}>0$ there exists, by Lusin’s theorem, a compact set $K \subset {\mathbb{I}}$ such that $A|_K$ is continuous and $m(K^c)<{\varepsilon}$. If $n$ is large enough then for every $x,y \in K$ that are $2^{-n}$-close we have $\|A(x)-A(y)\|<{\varepsilon}$. Let $G_n 