Bonjour to everybody.

I have to explain some notations before asking a simple question quoted from my favorite exercise book. Sorry about that.

First of all $\mathbb R$ is the set of real numbers. Use $z$ to denote a complex number, with real part $\Re z=x$ and imaginary part $\Im z=y$, so $z=x+iy$.

Let us denote by $\mathbb H$ the set of all complex numbers with strictly positive imaginary part, so $z=x+iy \in \mathbb H$ implies $y>0$. Let $\mathbb H'$ denote $\mathbb H\setminus\{i\}$.

We set $z=x+iy\in\mathbb H$, and after some calculations we show that the imaginary part of $\frac{z\cos\theta-\sin\theta}{z\sin\theta+\cos\theta}$ is strictly positive: $$\Im(\frac{z\cos\theta-\sin\theta}{z\sin\theta+\cos\theta})=\frac{y}{(x\sin\theta+\cos\theta)^2+(y\sin\theta)^2}>0,$$ and consequently $$\frac{z\cos\theta-\sin\theta}{z\sin\theta+\cos\theta}\in \mathbb H.$$

Now we write, for all $z$ in $\mathbb H$ and any real $\theta$, $$A_\theta(z)=\frac{z\cos\theta-\sin\theta}{z\sin\theta+\cos\theta}.$$

We now define a function $U:\,\,]0,1[\,\times\mathbb R\to\mathbb H$: $$U(t,\theta)=A_\theta(it)=\frac{it\cos\theta-\sin\theta}{it\sin\theta+\cos\theta}.$$

$U$ is infinitely differentiable and $\pi$ periodic with respect to $\theta$. In symbols, $U\in C^\infty(]0,1[\times\mathbb R)_{per}$.

We want to build an isomorphism $V$ by associating to any function $\varphi \in C^{\infty}(\mathbb H')$, the function $\psi=\varphi \circ U$: $$V:C^{\infty}(\mathbb H') \rightarrow C^{\infty}(]0,1[ \times \mathbb R)_{per},$$ $\varphi \mapsto \psi=\varphi \circ U$.

We first notice that the application $\varphi \circ U$ is a composition of $C^{\infty}$ class functions so $V$ is $C^{\infty}$ too.

$V$ is a linear application and $V$ is also $\pi$ periodic with respect to $\theta$.

In order to show that $V$ is an isomorphism, we have to determine the inverse isomorphism $V^{-1}$:

$\psi \mapsto \varphi=\psi \circ U^{-1}$

But I thought that the inverse of a composition function $$\forall x \in X,(g \circ f)(x)=g(f(x))$$ was defined by: $$(g \circ f)^{-1}=f^{-1} \circ g^{-1}.$$

In that case, can someone tell me

Why is the inverse isomorphism $V^{-1}$ not equal to $(\varphi \circ U)^{-1}=(U^{-1} \circ \varphi^{-1})$ instead of $\psi \mapsto \varphi=\psi \circ U^{-1}$ ? (knowing that a composition of functions is not commutative...)

More simply,

How can I write an explicite formulation of an inverse isomorphism of a composition of functions.

I thank you in advance for your help.


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  • $\begingroup$ Two notational questions: Is $\mathbb H'$ the same as $\mathbb H$? Is the subindex ${}_{per}$ intended to convey we are extending the functions periodically on $\theta$ with period $\pi$? $\endgroup$ – Andrés E. Caicedo Jul 6 '13 at 17:34
  • $\begingroup$ Oh i forgot that: H' is equal to H less the number {i}. $\endgroup$ – user43414 Jul 6 '13 at 18:16
  • $\begingroup$ and subindex "per" qualifies the set of infinitely differentiable functions (like U) that are periodic with a period of "\pi" with respect to $\theta$... $\endgroup$ – user43414 Jul 6 '13 at 18:29
  • $\begingroup$ Thank you. I edited the question to include this in the body of the text. $\endgroup$ – Andrés E. Caicedo Jul 6 '13 at 19:56

You are confusing two types of maps. We have functions $\varphi$ and $\psi$ defined on $\mathbb{H}$ and $(0, 1)\times\mathbb{R}$ respectively, whereas $V$ and $V^{-1}$ are maps between sets of such functions.

\begin{align*} V : C^{\infty}(H') &\to C^{\infty}((0, 1)\times\mathbb{R})\\ \varphi &\mapsto \varphi\circ U \end{align*}

That is $V(\varphi) = \varphi\circ U$. Therefore $V^{-1}$, the inverse of $V$, satisfies $V^{-1}(\varphi\circ U) = \varphi$ so

\begin{align*} V^{-1} : C^{\infty}((0, 1)\times\mathbb{R}) &\to C^{\infty}(H')\\ \psi &\mapsto \psi\circ U^{-1} \end{align*}

is the inverse of $V$. Check: $V^{-1}(V(\varphi)) = V^{-1}(\varphi\circ U) = (\varphi\circ U)\circ U^{-1} = \varphi$ and likewise for $V(V^{-1}(\psi))$.

As $\varphi\circ U : (0, 1)\times\mathbb{R} \to \mathbb{H}$, the map $(\varphi\circ U)^{-1}$ (if it exists) would be a map $\mathbb{H} \to (0,1)\times\mathbb{R}$. This has nothing to do with $V$.

  • $\begingroup$ to Michael Albanese: thanks a lot ! I had to reason in terms of maps between sets. This is now clear to me. $\endgroup$ – user43414 Jul 6 '13 at 18:41

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