# Differential of a smooth extension of a map

Let $X\subset R^N$, $Y\subset R^M$ be manifolds in the sense of Guillemin-Pollack (every point in $x\in X$ has a neighborhood locally diffeomorphic to an open subset of $R^k$, and similarly for $Y$). A map $f: X\rightarrow Y$ is smooth if it extends to a smooth map $F: U\rightarrow R^M$ where $X\subset U$. ($F$ must agree with $f$ on $X$.)

Probably this is a silly question, but is it true that $dF_x=df_x$? I don't see directly from definitions why this must hold. The definition of $df_x$ is as follows:

If $\phi: V\rightarrow X, x'\rightarrow x, \psi: W\rightarrow Y, y'\mapsto y$ are local parametrizations around $x$ and $y$ and if $h=\psi^{-1}\circ f\circ \phi$, then $df_x=d\psi_{y'}\circ dh_{x'}\circ d(\phi^{-1})_x$

• Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – GNUSupporter 8964民主女神 地下教會 Jan 23 '18 at 22:26
• I fixed the question. – user500094 Jan 23 '18 at 22:33
• +1 for your effort. I observe some changes in the notations from the picture. I believe its logically correct though. You may want to know how to typeset commutative diagram with MathJax in the tutorial. – GNUSupporter 8964民主女神 地下教會 Jan 23 '18 at 22:41
• Just an important comment. I think that what you meant to write as the definition of the differential was $df_x = d\psi_{y'}\circ dh_{x'} \circ d\phi^{-1}_{x'}$, following Guillemin-Pollack, where they consider $y'=\psi(0)$,$x'=\phi(0)$ and then define $df_x = d\psi_0 \circ dh_0 \circ d\phi^{-1}_0$. So in the definition we have the inverse of the differential of $\phi$ at the point $0$, and not the differential of the inverse $\phi^{-1}$ at the point $x$. – Edmundo Martins Jan 24 '18 at 0:53
• @EdmundoMartins Thanks for the remark. I thought $(d\phi_{x'})^{-1}=d(\phi^{-1})_x$ since $\phi$ is an diffeomorphism of $V$ onto $\phi(V)$. Is this true? – user500094 Jan 24 '18 at 4:31

They are not exactly equal, as $dF_x$ is defined on $\mathbb{R}^N$ and $df_x$ is defined only on the tangent space of $X$ at the point $x$, which I believe Guillemin and Pollack denote by $T_x(X)$.
But if you restrict $dF_x$ to $T_x(X)$, then the maps are the same. To see this, notice that as $F$ is a local smooth extension of $f$ at the point $p$, it is defined on an open subset $W \subset \mathbb{R}^N$ containing $x$ such that $F$ agrees with $f$ along $W \cap M$. Note that $W \cap M$ is an open subset of $M$ (we are considering the induced topology on $M$), so by taking $V$ and $U$ sufficiently small, we may assume that $\phi(U) \subset W \cap M$ and that the composition $\psi^{-1} \circ F \circ \phi: U \to V$ is well-defined and is actually equal to $h = \psi^{-1} \circ f \circ \phi$ (I am using the same notation as you did in your question).
Now, if we want to apply $dF_x$ to a vector $v \in T_x(X)$, by yhe definition of the tangent space we have that $v = d\phi_{x'}(z)$ for some $z \in U$. Then $$dF_x(v) = dF_x(d\phi_{x'}(z)) = d(F \circ \phi)_{x'}(z),$$ where I have used the usual Chain Rule for maps between open subsets of the Euclidean space above. But we have the following commutative diagram: $$\require{AMScd} \begin{CD} W \cap M @>F>> Y\\ @A \phi AA @A\psi AA\\ U @>h>> V \end{CD}$$ from which we can conclude that $F \circ \phi = \psi \circ h$. Taking differentials at the point $x'$, evaluating at $z$ and using the Chain Rule once again yields $$d(F \circ g)_{x'}(z) = d(\psi \circ h)_{x'}(z) = d\psi_{y'} \circ dh_{x'}(z).$$ Now since $v=d\phi_{x'}(z)$, we have $v=(d\phi_{x'})^{-1}(v)$. Pluggin this in the equation above gives us $$dF_x(v) = d(F \circ g)_{x'}(z) = d\psi_{y'} \circ dh_{x'} \circ d\phi^{-1}_{x'}(v) = df_x(v),$$ so that $dF_x(v)=df_x(v)$ in the sense of the defintion given.
In resume, the differentials $df_x$ and $dF_x$ are equal if we restrict $dF_x$ to $T_x(X)$. In Milnor's book Topology from the Differential Viewpoint, the differential of a smooth map $f$ between manifolds is actually defined as the differential of any smooth local extension $F$.