# Inverse Function Theorem as stated in Guillemin and Pollack

The following is an excerpt from Differential Topology by Guillemin and Pollack

But $f$ is certainly not the identity as stated above, so what exactly do G&P mean when the say "then one can choose local co-ordinates around $x$ and $y$ so that $f$ appears to be the identity"?

From what I gather from the commutative diagram, this 'reformulation' of the Inverse Function Theorem states that for $x \in X$, if $df_x$ is an isomorphism, then for parameterizations $\phi : U \to X$ and $\psi : U \to Y$ we have $f = \psi \circ \phi^{-1}$

But even if we have $f = \psi \circ \phi^{-1}$, $f$ need not be the idenity function. So what exactly were G&P trying to say here?

• What G&P are trying to say here is that you can understand your manifolds in terms of some atlas that you put on it. You can choose your charts to be such that $\psi^{-1} \circ f \circ \phi$ is the identity (as shown in the diagram). – Maxime Scott Jul 28 '17 at 19:21
• They mean exactly what they write starting with the words "...there exist local parameterizations..." – Lee Mosher Jul 29 '17 at 0:23

The Inverse Function Theorem that if $f:X\to Y$ is a smooth map and $df_x$ is an isomorphism, then $f$ is a local diffeomorphism at $x.$ Hence that for some $W$ open in $X,$ the restriction $f|_W$ is a diffeomorphism onto its image $f(W).$ If necessarily, we can shrink $W$ so that there exist local coordinates $\phi:U \to W$ and $\psi:U \to f(W).$ Then $f=\psi \circ \text{id}_U \circ \phi^{-1}.$
Now, on $U$ we have the coordinates $\phi^{-1}=(x_1,\ldots,x_k)$ and the coordinates $\psi^{-1}=(y_1,\ldots,y_k).$ Let $w \in W.$ Then $w=\phi(u)$ for some $u \in U.$ In coordinates we have $$u=(x_1(w),\ldots,x_k(w))=(y_1(w'),\ldots,y_k(w'))$$ for some $w' \in f(W).$ When it says that $f$ "appears to be the identity" it refers to the fact that $$f(w)=f(\phi(u))=\psi(u)=w',$$ and $w,w'$ have the same "representative" element $u \in U.$ That's why usually, in abuse of notation, we write the coordinates $(y_1,\ldots,y_k)$ in $f(W)$ as the same coordinates $(x_1,\ldots,x_k)$ in $W,$ and write $f(x_1,\ldots,x_k)=(x_1,\ldots,x_k),$ when it (strictly) should be $f(x_1,\ldots,x_k)=(y_1,\ldots,y_k).$
• @RagingBull Take $W\subset X$ such that $f|_W:W\to f(W)$ is a diffeomorphism. Since $X$ is a manifold there is local coordinates $\phi:U_1\to W_1$ where $x\in W_1,$ and similarly there is local coordinates $\psi:U_2\to W_2$ where $f(x)\in W_2.$ Using translations, we may assume $0\in U_1,$ $\varphi(0)=x,$ and $0\in U_2,$ $\psi(0)=f(x).$ Then let $U=U_1\cap U_2\cap \varphi^{-1}(W\cap W_1)\cap \psi^{-1}(f(W)\cap W_2),$ which is an open neighborhood of $0.$ Then change $W$ by $\varphi(U)\cap W$ and $f(W)$ by $\psi(U)\cap f(W),$ and restrict your homeomorphisms $\varphi$ and $\psi.$ – positrón0802 Mar 7 at 1:00
$f$ is not the identity of course... it's just a map from $X$ to $Y$ with certain properties. What they are trying to say is that you can choose coordinates for some neighborhood of $x$ and its image so that $f$ maps the coordinate grids identically on top of one another.