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I am starting to read the book "Differential Forms in Algebraic Topology" by Bott and Tu.

In the proof of the exactness of the Mayer - Vietoris sequence (Proposition 2.3, page 22 - 23) a partition of unity $\{\rho_U,\rho_V\}$ subordinate to an open cover of two open sets $U,V$ is applied to a function $f$ which is defined on the intersection $U \cap V$.

Then the authors emphasize that $\rho_V \,f$ is a function on $U$. I struggle to see why this is true, I thought $\rho_V$ has support contained in $V$, and so from the picture in the book I have the impression that the function $\rho_U\,f$ is a function with support in $U$. I am aware this must be false, the authors even stress the fact that one has to multiply by the partition function of the other open set! Any help to understand this would be great, many thanks !!

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    $\begingroup$ "The authors". What authors? What book? What source? What is $\rho_Vf$? $\endgroup$ – Arturo Magidin Jul 7 '12 at 20:56
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    $\begingroup$ the book I am referring to is titled "Differential Forms in Algebraic Topology" written by R. Bott and L. Tu, and $p_V \,f$ is the product of a smooth function defined on the intersection of two opens subsets $U,V$ of $\mathbb{R}$ with a partition function $\rho_v$ whose support is contained in $V$. $\endgroup$ – harlekin Jul 7 '12 at 21:04
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    $\begingroup$ @Arturo: It seems to me that the only information in harlekin's comment that is not in the post already is that $f$ is a smooth function and that $U$ and $V$ are subsets of $\mathbb{R}$, neither of which is essential to the question. Perhaps you're being a bit harsh? $\endgroup$ – Zev Chonoles Jul 7 '12 at 21:10
  • $\begingroup$ @Zev: I could swear the title of the book was not there when I posted my original comment... Given the timing, it's possible either I missed it or that it was added within the 5 minute winddow. $\endgroup$ – Arturo Magidin Jul 7 '12 at 21:16
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The support (Wikipedia article) of $\rho_V$ is contained in $V$, so that $\rho_V(x)=0$ for any $x\notin V$. Thus, even though $f$ is only defined on $U\cap V$, we can extend $\rho_Vf$ to $U\cap V^c$ by declaring $\rho_Vf(x)=0$ for all $x\in U\cap V^c$ (this extending of the domain of the function is being done implicitly by the authors). This defines $\rho_Vf$ on all of $U=(U\cap V)\cup (U\cap V^c)$.

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