Consider $M$ a smooth compact connected manifold with $w$ a volum form. Take for example $M=\mathbb{S}^n$ with the uniform measure. For any smooth map $f:M\rightarrow M$ and its pullback measure $\nu = f^* w$ it is well known that we have $$\int_M \nu = \text{deg}(f)\int_M w $$with $\deg(f)\in \mathbb{Z}$. Is this condition sufficient? For any $\nu$ volum form on $M$ such that $\int_M \nu / \int_M w \in \mathbb{Z} $ does there exist $f:M\rightarrow M$ such that $\nu = f^*w$ ? Can we construct such an $f$ explicitly ?
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4$\begingroup$ $S^n$ does not have a Haar measure unless $n=0,1,3$. Maybe $7$ if you stretch the meaning of Haar measure a bit. Ignoring this, most manifolds do not have self-maps of arbitrary degree; and if you want $\nu$ to also be nonvanishing then $f$ should be a covering map (even less common). It is a theorem of Moser that if $\int \omega = \int \nu$ for two nonvanishing top forms, then there's an oriented diffeomorphism $f$ with $f^*\nu = \omega$. The proof is what's usually called the Moser trick. $\endgroup$– mmeCommented Sep 3, 2020 at 13:09
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$\begingroup$ a note on terminology: usually measures are only pushed forward, and differential forms are pulled back. Defining pullback measures is tricky mathoverflow.net/q/122704/3948 . $\endgroup$– Willie WongCommented Sep 3, 2020 at 16:08
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No. Consider the $2$-sphere, and the standard volume $\omega$ form with volume $4\pi.$ Consider $\nu=2\omega.$ If $f\colon S^2\to S^2$ with $f^*\omega=\nu$ would exist, it would be a local diffeomorphism and by compactness a covering map. As $S^2$ is simply connected and $f$ would have degree 2, this is not possible.
