How to prove that a function is not injective How can I prove that $f(x) = 1-x^2$ is not a one-to-one (injective) function?
 A: Let $f:\mathbb{R}\to\mathbb{R},x\mapsto 1-x^2$.
We have $-1\neq 1$ and $f(-1)=f(1)$. This proves that $f$ is not injective.
More generally, if $f:X\to Y$ is a map. Saying that $f$ is not injective is equivalent to the existence of two distinct elements $x,x'\in X$ such that $f(x)=f(x')$.
Warning : for a map $f:\mathbb{R}\to\mathbb{R}$, strict monotonicity implies injectivity (and hence non injectivity implies that $f$ is not strict monotonic), but the converse is false : for example the map defined by
$$f:\mathbb{R}\to\mathbb{R},x\mapsto\cases{1/x\quad\mathrm{if}\,x\neq0\cr0\quad\mathrm{otherwise}}$$is injective (it is a bijection and even an involution), but it is not monotonic.
EDIT
The OP asked how to prove "the non-injectivity of $f(x)=1-x^2$". This should be rephrased as : 

How to prove the injectivity of $f:\mathbb{R}\to\mathbb{R},x\mapsto1-x^2$

Here is why ...
The domain is an attribute of an application (and the same for the range) since any application is defined as a triple $(A,B,\Gamma)$ where the graph $\Gamma$ is a subset of $A\times B$, with the additional condition that every element in $A$ has a unique image in $B$.
For example, be saying $f:\mathbb{R}\to\mathbb{R}$ is (or is not) injective, the domain is $\mathbb{R}$.
If I want to change it to some subset $D$ of $\mathbb{R}$, I have to replace $f$ by its restriction to $D$ (which is then another application), and this is the meaning of "$f$ is injective on $D$" which is an abusive formulation for "$f$ restricted to $D$ is injective".

The important point is that an application $f:A\to B$ which is not injective may have an injective restriction to some $D\subset A$ :

$f:\mathbb{R}\to\mathbb{R},x\mapsto1-x^2$ is not injective, but
$g:[0,+\infty)\to\mathbb{R},x\mapsto1-x^2$ is (strictly decreasing hence) injective
A: You can consider that $x=3$ and $x=-3$ give $f(3)=f(-3)$ while $3 \neq -3$
A: Can an even function be injective?
