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The question is very vague therefore any kind of suggestions, reference, ideas are welcome.

Suppose $S$ is an oriented surface with or without boundary. Let $m$ be an area form. Let $f$ be a diffeomorphism of $S$ isotopic to the identity. Is there a way to decompose $f$ naturally such that one component preserves $m$. By naturally I mean the decomposition is continuous with respect to some topology on the group of diffeomorphisms isotopic to the identity.

(over) Expectation: There exist $A,B\subset S$ such that $A\cup B=S$, $m(A)\neq 0,$ (also $m(B)\neq 0$ if possible) and $f_1=f_{|A}, f_2=f_{|B}$ such that $f_1$ preserves $m$.

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    $\begingroup$ This is really too vague to answer. What do you mean by "decompose"? What do you mean by "component"? And what operation do you mean by $+$? $\endgroup$
    – Andy Putman
    Commented Apr 23, 2018 at 16:26
  • $\begingroup$ @AndyPutman sorry for being too much vague. I have edited the question. $\endgroup$
    – Cusp
    Commented Apr 23, 2018 at 17:00
  • $\begingroup$ What makes you think this holds even for a single diffeomorphism? $\endgroup$
    – Andy Putman
    Commented Apr 23, 2018 at 17:22

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If the Jacobian of $f$ is not equal $1$ on any set of positive measure, then the mapping will not be measure preserving on any set of positive measure so a decomposition is not possible.

A classical result of Moser (see e.g. Lower regularity version of Moser's theorem on volume elements) states that for any smooth positive function $g$ on $S$ whose integral equals to the area of $S$ there is a diffeomorphism with the Jacobain equal to $g$ and clearly we can take $g$ to be equal $1$ only on a set of measure zero (on a curve).

I am not sure what the author means by preserving $m$ so I took a freedom to understand it as a measure preserving map: https://www.encyclopediaofmath.org/index.php/Measure-preserving_transformation

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