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Here I am using de Rham cohomology. This question occured to me while reading the proof of the exactness of the short exact sequence in the Meyers Vitoris sequence

$0 \rightarrow H^{n}(X)\,\xrightarrow{i}\,H^{n}(U)\oplus H^{n}(V)\,\xrightarrow{j}\,H^{n}(U\cap V)\rightarrow 0$

Where $j$ is defined by $j(\theta, \tau)=\theta - \tau$. In Bott and Tu specifically (and other proofs I looked at trying to see what was happening) to prove the surjectivity of $j$ take a closed differential form on $U\cap V$, $\omega$ lets say, and then take a partition of unity $\rho_U, \rho_V$ of $X$. Then $(\rho_U \omega , -\rho_V \omega)$ should be the element in $H^{n}(U)\oplus H^{n}(V)$ mapping to $\omega$. However it is not clear to me at all that $\rho_U \omega$ is a well defined smooth form on $U$, because there are, for example, functions smooth on subsets of $\mathbb{R}$ that cannot be extended smoothly to full space e.g. $tan(\theta), x^{-1}$ and so on. So is it different for closed forms?? Can a closed differential form on a subset of manifold always be extended to the whole manifold?

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    $\begingroup$ "Can a closed differential form on a subset of manifold always be extended to the whole manifold?" No: e.g. $tan(\theta)d\theta, x^{-1}dx $ :-) $\endgroup$ – Georges Elencwajg Feb 18 '16 at 18:05
  • $\begingroup$ "The right question" (which will remove @Georges's objection) is to ask whether forms on a closed submanifold can be so extended. (This is not the context of the Mayer-Vietoris setting, but turns the question in a slightly different direction.) $\endgroup$ – Ted Shifrin Feb 20 '16 at 5:41
  • $\begingroup$ @Ted thank you for your comment, I think I'm still quite hazy on the subtleties of everything. I am happy to change the question to the closed submanifold case, I can see that this is a harder question which I haven't thought about at all, but I don't think it answers the original problem I have with the proof of the Meyer Vietoris sequence. I don't think there was something that indicates $U \cap V$ has to be closed, I will check more thoroughly $\endgroup$ – R Mary Feb 20 '16 at 14:31
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Your sequence is not correct I think, these should not be the cohomology, but the spaces of forms.

Note that $\rho_U$ has support in $U$. Hence there is a closed set $C\subset U$ such that $\rho_U(x)=0$ for all $x\in U\setminus C$.

Thus one can define a form

$$ \rho_U \omega=\begin{cases} \rho_U \omega\qquad \text{on } & U\\ 0\qquad\text{not on} & U. \end{cases} $$

This is smooth.

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  • $\begingroup$ Thank you for your comment but Im afraid I don't see that the form so defined is smooth. Following above take $U,V$ to be subsets of $\mathbb{R}$ so that $U=\{x \in \mathbb{R} | x > -\frac{\pi}{2}}$ and $U=\{x \in \mathbb{R} | x < \frac{\pi}{2}}$. Then the form $\omega=tan(x)dx$ is smooth on $U \cap V$ but $\rho_U \omega$ does not seem to even exist on $U$. What am I misunderstanding?? $\endgroup$ – R Mary Feb 20 '16 at 14:13
  • $\begingroup$ You're correct. The form $\rho_U\omega$ is globally defined on $V$, but not on $U$. $\endgroup$ – Ted Shifrin Feb 20 '16 at 17:59
  • $\begingroup$ @R Mary: Let's do this for the continuous case, the smooth case is similar but writing down explicit functions becomes a mess. Let $f(x)=1-x^2$ for $|x|\leq 1$ and $x=0$ for $x\geq 1$. This is a continuous function. Set $g(x)=1-f(x)$. Then $f+g=1$ everywhere, and $f$ has support in $U=(-2,2)$. Set $V=\mathbb{R}\setminus [-1,1]$ Let $\omega$ be $\frac{1}{4-x^2} dx$. Clearly $\omega$ does not extend to $U\cup V=\mathbb{R}$. But the form $f\omega$ is can be seen as continuous form on $\mathbb{R}$ (by setting it to be $0$ outside of $U$). I think this is where your confusion lies. $\endgroup$ – Thomas Rot Feb 21 '16 at 11:32

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