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I am just verifying,

Is the function $f(x) = \tan(x)$ onto but not one to one?

tangent

Tsangares
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    Yes. You have it exactly -- it's not one to one because it fails the horizontal line test (or because it is periodic). See http://math.stackexchange.com/a/1178141/98077 for a proof that it is onto. – walkar Feb 08 '17 at 05:28
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    Only in a period you can say it's 1-1. – Nosrati Feb 08 '17 at 05:30

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Yes. It is onto the set $\mathbb R\cup\{\infty\},$ where $\text{“}\infty\text{''}$ is neither $+\infty$ nor $-\infty$ but rather the single $\text{“}\infty\text{''}$ that is approached by going either in the positive or in the negative direction.

It is locally one-to-one at every point. That means that about every point $x$ in its domain there is some open interval $(x-\varepsilon,x+\varepsilon)$ such that the restriction of the tangent function to that interval is one-to-one.

But the tangent function is not one-to-one. There are many points $x$ in its domain for which $\tan x = 1.$ And many for which $\tan x=2,$ etc.

By contrast, the sine function is not locally one-to-one at $\pi/2.$ No matter how small you make $\varepsilon$, the restriction of the sine function to the interval $(\pi/2-\varepsilon,\pi/2+\varepsilon)$ is not one-to-one.

Michael Hardy
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