Verification of my work in injectivity and surjectivity of f(x)=1+x^2

Ozma

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Verify that [imath]f^{-1} [/imath] exists for [imath]f:\mathbb{R}\to\mathbb\{x\in\mathbb{R} \ | \ x>1\}[/imath] defined by [imath]f(x)=1+x^2[/imath].

For the surjectivity: let [imath]y>1[/imath], it is [imath]y=1+x^2 \iff x^2=y-1 \iff x=\sqrt{y-1} \ \text{or} \ x=-\sqrt{y-1}[/imath]. In both cases [imath]\sqrt{y-1}[/imath] and [imath]-\sqrt{y-1}[/imath] are real numbers for [imath]y>1[/imath], hence they are acceptable values.
Since [imath]y>1[/imath] is generic, this holds for any [imath]y>1[/imath] and so [imath]f[/imath] is surjective.

However, [imath]f[/imath] is not injective because [imath]f(1)=2[/imath] and [imath]f(-1)=2[/imath]; my textbook (Nicholson - Introduction to Abstract Algebra, 4th Edition, Exercise 0.3.7 b) says that I should verify that [imath]f[/imath] is bijective and find [imath]f^{-1}[/imath] (the text of the problem is exactly what I wrote in the beginning), but I don't see any mistakes in my study. Is my textbook wrong or am I doing something wrong? Or with "verify" the author doesn't mean that [imath]f^{-1}[/imath] surely exists and I have to determine whether it exists or not?
 
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I was confused by the problem's statement at first since it uses [imath]x[/imath] both for the domain and for the image of [imath]f[/imath]. But your answer looks good to me.
 
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