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Let $M$ be a compact manifold and $\varphi:C^\infty(M)\rightarrow \mathbb{R}$ be a function which assigns to every $f\in C^\infty(M)$ the value $\int_M fdV.$

In a smooth topos which is a well adapted model, does the morphism $\overline{\varphi}:R^M\rightarrow R$ corresponding to $\varphi$ exist?

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  • $\begingroup$ Say more about the volume measure...is it given with the manifold? Does it come from a metric? or from an embedding into R^n? $\endgroup$
    – user44143
    Commented Feb 13, 2017 at 14:23
  • $\begingroup$ I'm especially interested in the case it comes from a metric, but if there is a discussion for other cases, I want to know that. $\endgroup$
    – user80247
    Commented Feb 13, 2017 at 16:10
  • $\begingroup$ The answer seems to be yes for Riemannian M, using Anders Kock's construction of a "volume form as volume of infinitesimal simplices", arxiv.org/abs/math/0006008 $\endgroup$
    – user44143
    Commented Feb 13, 2017 at 18:19
  • $\begingroup$ I find it hard to believe that it matters how the volume form is given, since by Moser's theorem any two volume forms on a compact manifold are equivalent up to a diffeomorphism . $\endgroup$
    – Michael Bächtold
    Commented Feb 15, 2017 at 8:08
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    $\begingroup$ @MattF. I skimmed Kock's article but couldn't see the result the OP wants there. Could you be more specific about how it follows from the results in the article? $\endgroup$
    – Michael Bächtold
    Commented Feb 15, 2017 at 8:11

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