# What do we mean when we say a differential form “descends to the quotient”?

Let $S$ be a surface and let $f:S\rightarrow S$ be a diffeomorphism. We define the mapping torus $M_f$ of the pair $(S,f)$ to be the quotient

$$(S\times I) /\sim \quad \text{ where } \ (1,x) \sim (0,f(x))$$

i.e., we "glue the cylinder"$S\times I$ along $f$"

I'm given a contact differential form $w$ defined on $S$, and I'm asked to check that the form $w$ descends to a contact form on the mapping torus of $(S,f)$.

Now I know that a necessary condition for any function $f$ to pass to the quotient given a quotient map $q$ is for the function to be constant on quotient classes of $q$, but, do I need some other condition, say, on the pullback of $w$ by the quotient map $q:S\rightarrow M_f$?

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Is $w$ really defined in a surface (2-dimensional?). If so, how can it be a contact form? – 40 votes Jul 21 '13 at 6:01

on the pullback of $w$ by the quotient map $q:S→M_f$
First, the quotient map is from $S\times I$ to $M_f$, not from $S$. Second, you can't pull back $w$ by the quotient map because $w$ lives on the domain of the quotient map.
What you want to show is that there is a differential form $\zeta$ on $M_f$ such that the pullback of $\zeta$ by $q$ is $w$.