Vtyjkyuk

In "Loring W. Tu, An introduction to manifolds" the following question exists: Let $q$ be a point of an $n$-dimensional manifold $M$ and $U$ any neighborhood of $q$. Construct a smooth bump function at $q$ supported in $U$. I answered that question but I want to make sure. Here is my answer:

Let $q$ be arbitrary of $M$ that is contained in a neighborhood $Usubset M$. Then, there exists a coordinate chart $(V,phi)$ in the maximal atlas of $M$ such that $qin V subset U$. In particular, there exists a smooth bump function $rho:mathbbR^n to mathbbR$ at $phi(q)$ supported in $phi(V)$ that is identically $1$ in a neighborhood $B_r(phi(q)) subset phi(V)$, say, of $phi(q)$. Define a map $f:Mto mathbbR$ by
$$
f(p) =
begincases
rhobigl(phi(p) bigr), &text$pin V$, \
0, &text$pnotin V$.
endcases
$$
Being the composite of two smooth function, $f$ is smooth on $V$ and hence on the whole manifold $M$. If $pin phi^-1bigl( B_r(phi(q)) bigr)$, then $phi(p) in B_rbigl( phi(q) bigr)$ and therefore, by the construction of $rho$, $rhobigl( phi(p) bigr)=1$. That is, $f equiv 1$ on the neighborhood $phi^-1bigl( B_r(phi(q)) bigr)$ of $q$. Clearly, by the definition of $f$, $supp, f subset V subset U$. Hence, $f$ is a smooth bump function at $q$ supported in $U$.

Can anyone please revise my proof ?. I appreciate your help.
Thanks in advance.

Note: A smooth bump function $f:Mto mathbbR$ at a point $qin M$ supported in $Usubset M$ is a non-negative smooth function such that $fequiv 1$ on a neighborhood $V_q subset U$ of $q$ and that $supp, f subset U$.

As pointed out in the comments, the map $f$ that you constructed may not be $C^infty$. But you are almost there.

Notice that in Tu's book the bump function $phi:mathbbR^nrightarrow mathbb R$ could be chosen so that its support $mathrmsuppphi$ is a compact subspace of $mathbbR^n$. Then $phi^-1(mathrmsupprho)$, being the image of a compact subspace under the continuous map $phi^-1:phi(V)rightarrow V$, is a compact subspace of $V$. Hence $phi^-1(mathrmsupprho)$ is a compact subspace of $M$. But $M$ is Hausdorff, so this means that $phi^-1(mathrmsupprho)$ is closed in $M$. Therefore, we have that $$mathrmsuppf=mathrmcl_M((rhocircphi)^-1(mathbbR^times))=mathrmcl_M(phi^-1(rho^-1(mathbbR^times))subsetmathrmcl_M(phi^-1(mathrmsupprho))=phi^-1(mathrmsupprho)subset V.$$
From this follows that $f$ is smooth: If $pnotin V$, choose a chart about $p$ disjoint from $phi^-1(mathrmsupprho)$. (This is possible because $phi^-1(mathrmsupprho)$ is closed in $M$.) Then this chart is also disjoint from $mathrmsuppf$, so on this chart $f$ is identically $0$. Hence $f$ is $C^infty$
at $p$. This proves that $f$ is $C^infty$ at every point not in $V$. By construction, $f$ is also $C^infty$ at every point in $V$. Hence $f$ is $C^infty$.

Finally you check that $f$ satisfies all the properties required as a bump function at $q$ supported in $U$. This should not be difficult.