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In the GTM218,Theorem 18.7:

For any smooth manifold $M$,the map $l_*:H_p^{\infty}(M)\rightarrow H_p(M)$ induced by inclusion is an isomorphism.

Is it also true for smooth manifold with boundary?(It may be wrong because it uses the Whitney approximation theorem) I want to know this because I want to define the de Rham homomorphism on smooth manifold with boundary,but the book only deals with smooth manifold without boundary.(And Stokes' theorem for chains only deals with smooth manifold without boundary too.)

I think that's right, otherwise it's disappointing.

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    $\begingroup$ This question and my answer to it are relevant. $\endgroup$
    – Jack Lee
    Mar 30, 2019 at 19:42
  • $\begingroup$ @JackLee Thanks.But actually I would like to know if $c,c^{'}\in C_1^{\infty}(M)$ are homologous in singular chain complex(not smooth singular chain complex),will they differ by a boundary in $\partial(C_2^{\infty}(M))$?($M$ is a smooth manifold with boundary.) $\endgroup$ Mar 31, 2019 at 7:33
  • $\begingroup$ The point is that the inclusion map from $C_*^{\infty}$ to $C_*$ is a chain-homotopy equivalence. You can either prove this directly by imitating the proof for manifolds without boundary or follow Jack Lee's hint and use a composition of several chain homotopy-equivalences. $\endgroup$ Apr 2, 2019 at 17:56
  • $\begingroup$ JackKee said:" One problem is that my proof of the isomorphism between smooth singular & singular cohomology uses the Whitney approx theorem for a map that's already smooth on a closed subset, and this doesn't generally work when the target has nonempty boundary." Prove this directly by imitating the proof for manifolds without boundary may not work? $\endgroup$ Apr 3, 2019 at 3:03

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