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Prove that the mapping $z\mapsto\bar z$ of $\Bbb C\to \Bbb C$ is an isomorfism of $\Bbb C$ to itself that punctually fixes $\Bbb R$.

That's not so hard to prove, if $\bar z=a-ib=x-iy=\bar w \Rightarrow a=x$ and $-b=-y\Rightarrow b=y\Rightarrow z=w$, so it is inyective. And for any $\bar z$ there exists a $z$ such that $z\mapsto\bar z$, so it is surjective. But I don't understand the part where it says that the mapping punctually fixes $\Bbb R$, what does that mean? What I interpret is that for any $r\in\Bbb R$ the mapping send it to itself, wich is true since $\bar r=r $, $\forall r\in\Bbb R$. Do I have to prove something else?

Ana Galois
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What you're asked to prove, among another things, as you said, is that $\forall r\in \Bbb R(\overline r=r)$.

The choice of words is to emphasize that you're not asked to prove that $\Bbb R=\{\overline r\colon r\in \Bbb R\}$. The first condition is stronger.

I've never seen punctually used that way. Much more common is pointwise.

Git Gud
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  • The original text where I got this is in spanish, and literally says: "...que fija puntualmente $\Bbb R$". The sentence that I wrote is my own translation, so probably is pointwise, like you say. – Ana Galois Nov 27 '13 at 17:58
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    @AnaGalois Yes, it is. Lost in translation... When it comes to math terminology it's a good idea to visit your own language's wikipedia page, locate the thing you're looking for and then switch to english. – Git Gud Nov 27 '13 at 18:00