Yahoo Answers: Answers and Comments for Prove that the copying f:R>[1,+∞), f(x)=x²+1 is surjection but not injection? [Mathematics]
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From Anonymous
enSG
Mon, 22 Oct 2007 01:24:28 +0000
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Yahoo Answers: Answers and Comments for Prove that the copying f:R>[1,+∞), f(x)=x²+1 is surjection but not injection? [Mathematics]
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From Scarlet Manuka: For y ≥ 1, √(y1) is in R and f(√(y1)) = y. (...
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Mon, 22 Oct 2007 01:40:05 +0000
For y ≥ 1, √(y1) is in R and f(√(y1)) = y. (You can derive √(y1) by setting x^2 + 1 = y.) So for every y ≥ 1 there is an x in R such that f(x) = y, i.e. f is a surjection.
However, for x in R, f(x) = f(x) and so f is not an injection (e.g. f(1) = f(1) but 1 ≠ 1). Note that in the first part we could equally well have chosen x = √(y1), which is a different number to √(y1) for all y > 1.

From Curt Monash: It's not an injection because, for example...
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Mon, 22 Oct 2007 02:45:48 +0000
It's not an injection because, for example, 1 and 1 are both mapped to the same thing (namely 2).
It's surjection because for any y, we can solve x^2+1 = y. Specifically, set x = the square root of (y1).