TITLE: Verify whether a function is injective
QUESTION [7 upvotes]: I'm trying to learn how to verify whether a function is injective. This is my function:
$g=e^{1-y^2}-1$
How should I proceed in veryfing whether it is injective or not? 
In my lecture, we verified using the definition by checking whether the function is always increasing or decreasing (we started with only some part of a function and continued to add the rest of the function while always verifying whether $f(x_1)<f(x_2)$ for all $x_1<x_2$).
However, with this function, the approach seems too difficult to do. Is there any way to verify whether it is injective or not?
Thanks

REPLY [3 votes]: The simplest  way to see if a function is injective is to set $g(y) = g(y')$ and see if that does or does not mean $y = y'$
If $e^{1-y^2} -1 = e^{1-y'^2} -1$ then
$e^{1-y^2} = e^{1-y'^2}$
$\ln e^{1-y^2} = \ln e^{1-y'^2}$
$1-y^2 = 1-y'^2$
$y^2 = y'^2$
$y = \pm y'$
and that does NOT mean $y = y'$.
$g(y) = e^{1-y^2} -1$ is not injective as for any $a \ne 0$ we have $a\ne -a$ and $g(a) = g(-a)$.
... 
Another way; the Calculus way is to take the derivative.  
If the direvative is always positive or always negative it is injective. (If so the function is monotonically increasing or decreasing-- which means if $x < y$ then $f(x) \ne f(y)$. Otherwise, if the function is continuous, if there are areas where the function is decreasing, where $f'(x) < 0$ and areas where it is increasing, $f'(x)>0$ then ... it doesn't pass the "horizontal line test". )
In this case $g'(y) = e^{1-y^2}*(-2y)$.  $e^{1-y^2} > 0$ but $-2y >0$ if $y < 0$ and $-2y < 0$ if $y> 0$ so it isn't injective.