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The wikipedia entry on the inverse function theorem includes a section on the constant rank theorem, and gives a formulation of it which I haven't seen in books:

A smooth map $f:M\to N$ of constant rank about a point $p$ admits open $U\subset M,V\subset N$ and diffeomorphisms $U\cong \mathrm T_pM,V\cong \mathrm T_{fp}N$ such that the obvious composite equals the derivative $\operatorname d_p\!F:\mathrm T_pM\to \mathrm T_{fp}N$. Then, the intuition is that about $p$, $f$ "looks like its derivative".

How is this equivalent to/deducible from the more common formulation which states a smooth map has a chart on which it's either an inclusion or a projection?

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    I don't see instantly how to prove it (except perhaps using the exponential map), but I like the formulation! – 2017-01-14
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    @JohnHughes: I think it's just the usual proof of the rank theorem, except not making linear coordinate changes to make the derivative map the $r\times r$ identity. But the rest of the diffeomorphisms remain. I think. – 2017-01-14

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