I am having trouble working with smooth functions defined on a manifold. Is the following line of reasoning valid or not?
Let $f$ be a smooth function defined on a manifold $M$ with a local maximum at point $p$. Let $F$ be a smooth function defined in a neighborhood $U$ of $p$ which agrees with $f$ on $M\cap U$, the restriction of the set $U$ to the manifold.
Suppose I want to prove something like $df_p = 0$. Can I simply state that $df_p = dF_p$ which must be zero because $F$ also has a local maximum at $p$?
My problem is that I am not sure that $F$ is guaranteed to exist and whether or not I can use the normal rules of calculus with $F$. If my reasoning is wrong, should I just attack this problem via coordinate charts?
Let $\varphi:W\to M\cap U$ be a parametrization of $M$ near $p$, $W$ open in $\mathbb R^d$, $p=\varphi(a)$. Then $f\circ\varphi:W\to\mathbb R$ has a local maximum at $a$ and $d_a(f\circ\varphi)\equiv0$ (critical point). Now we translate this to $M$. The tangent space of $M$ at $p$ is $T_pM=\text{Im}(d_a\varphi)$. Since $f|M\cap U=F|M\cap U$ we have: $$ 0\equiv d_a(f\circ\varphi)=d_a(F\circ\varphi)=d_pF\circ d_a\varphi, $$ which exactly means that $d_pF$ vanishes on the image of $d_a\varphi$, which is precisely $T_pM$. Hence: $$ d_pf=d_pF|T_pM=0. $$ This completes the argument. The condition $d_pF|T_pM\equiv0$ is just the famous Lagrange multipliers criterion for conditioned critical points. Note that $F$ needn't have a local maximum at $p$.
Summing up, $F$ does exists for $U$ small enough and the usual calculus rules work, but restricted to tangent spaces.