Let $\pi: (X,\mathcal{M},\nu) \to (Y,\mathcal{N},\eta)$ be a measurable map i.e. $\pi^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$. I want to define a map from $L^{\infty}(Y,\eta)$ to $L^{\infty}(X,\nu)$. The natural thing to do is the following:
Define a map $\tilde{\pi}:L^{\infty}(Y,\eta) \to L^{\infty}(X,\nu)$ by $$\tilde{\pi}(f)=f\circ\pi,f \in L^{\infty}(Y,\eta)$$ Recall that $$\|f\|_{\infty}=\inf\left\{\alpha \ge 0: \mu\left(\{x: |f(x)|>\alpha\right)=0\right\}$$
Claim-1: $\tilde{\pi}(f)$ is measurable.
Proof: Let's assume that $\tilde{\pi}(f)$ is real-valued, without any loss of generality. Observe that $\tilde{\pi}(f)^{-1}([a,\infty))=\pi^{-1}\left(f^{-1}([a,\infty)\right)$. Since $f$ is a measurable, $f^{-1}([a,\infty)) \in \mathcal{N}$. Since $\pi$ is a measurable map, $$\tilde{\pi}(f)^{-1}\left([a,\infty)\right)=\pi^{-1}\left(f^{-1}([a,\infty))\right) \in \mathcal{M}$$
Claim-2: $\tilde{\pi}(f) \in L^{\infty}(X,\nu)$
Proof: Observe that for a.e $x \in X$ , we have $|f(\pi(x))| \le \|f\|_{\infty} < \infty$.
At this point of time, I need one assumption that $\pi_{*}(\nu)=\eta$. I don't see how else to move forward without this assumption.
Let $E_{\alpha}=\{x\in X: |f(\pi(x))|\ge \alpha\}$. Let $F_{\alpha}=\{y \in Y: |f(y)|\ge \alpha\}$. Then, $E_{\alpha}\subseteq\pi^{-1}\left(F_{\alpha}\right)$. For $\alpha > \|f\|_{\infty}$, observe that $\eta(F_{\alpha})=0$. Since $\pi^{-1}(F_{\alpha}) \in \mathcal{N}$, we have that $\nu(\pi^{-1}(F_{\alpha}))=0$ and hence $\nu(E_{\alpha})=0$. Thus, $\|\tilde{\pi}(f)\|_{\infty} \le \|f\|_{\infty}< \infty$.
Please let me know if there is something wrong with this proof. Thanks for the help!!
The function $\tilde {\pi} $ is not well defined function unless you make an extra assumption on $\pi$. You need the condition $\nu (\pi ^{-1}(E))=0$ whenever $E \in \mathcal N$ and $\eta(E)=0$. You don't need the stronger assumption that $\eta =\nu \circ \pi^{-1}$.