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I was wondering whether the following commuting diagram property was enough to characterize the exterior derivative (although I'm suspicious. Is this answered somewhere in the literature? my question came about after posting a comment in What is the exterior derivative intuitively?)

Let $f$ be any smooth map $f:M\to N$ where $M$ and $N$ are smooth manifold and consider the map $f^*: A(N)\to A(M)$ between the spaces of smooth sections of the exterior form bundles $\Omega(N^*)$ and $\Omega(M^*)$, $A(N)$ and $A(M)$, associated to the manifolds $N$ and $M$. Then the exterior derivative is the unique map $d$ such that the following diagram commutes (and its properties follow from this characterization):

$\require{AMScd}$ \begin{CD} A(N) @>d>> A(N)\\ @V f^* V V @VV f^* V\\ A(M)@>>d> A(M) \end{CD}

Thanks.

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    $\begingroup$ I'm pretty sure that $d = 0$ makes your diagram commute...? $\endgroup$
    – Stefan Waldmann
    Commented Nov 19, 2020 at 18:25
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    $\begingroup$ This (and linked paper) contains what you're after I believe: math.stackexchange.com/questions/918949/… $\endgroup$
    – Paul Reynolds
    Commented Nov 19, 2020 at 18:49
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    $\begingroup$ The paper @PaulReynolds references: Palais - Natural operations on differential forms. $\endgroup$
    – LSpice
    Commented Nov 19, 2020 at 19:05
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    $\begingroup$ The identity also makes your diagram commute! I think you further want to require that $d$ is a derivation of degree $1$, and even then I think you can at best hope to recover $d$ up to a scalar. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 19, 2020 at 20:40
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    $\begingroup$ @PaulReynolds Definitely what I was thinking about, thank you! $\endgroup$
    – Rachid Atmai
    Commented Nov 19, 2020 at 22:54

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