How can I prove that $f(x) = 1-x^2$ is not a one-to-one (injective) function?

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    $\begingroup$ Proof by counterexample. Calling your function $f$, find two distinct points $s,t$ which satisfy $f(s)=f(t)$. $\endgroup$ – joeb Feb 19 '17 at 23:00
  • $\begingroup$ Note that whether a function $f$ is or is not injective depends, in part, on the domain of $x$. $\endgroup$ – Namaste Feb 19 '17 at 23:04

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.


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

  • $\begingroup$ I would be very happy if you make note in your answer that the behavior of a function depends on the domain of the function. E.g., suppose $x \in \mathbb R, x\geq 1$? $\endgroup$ – Namaste Feb 19 '17 at 23:15
  • $\begingroup$ Very nice answer. +1 $\endgroup$ – DonAntonio Feb 19 '17 at 23:40
  • $\begingroup$ @amWhy: I agree with that ! Editing ... $\endgroup$ – Adren Feb 20 '17 at 5:30

You can consider that $x=3$ and $x=-3$ give $f(3)=f(-3)$ while $3 \neq -3$

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    $\begingroup$ And can you also mention how injectivity of a function depends, in part, on the domain of the function? $\endgroup$ – Namaste Feb 19 '17 at 23:16

Can an even function be injective?

  • $\begingroup$ Really? Methinks this at best a comment. Six words, all suffixed with a question mark. What this shows is that your "sense" of what constitutes a helpful hint is way off the mark. $\endgroup$ – Namaste Feb 19 '17 at 23:12
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    $\begingroup$ It is a hint, not a proof. Food for thought. That's the way I was taught, and I persist to think it's more stimulating, if you don't mind. $\endgroup$ – Bernard Feb 19 '17 at 23:21
  • $\begingroup$ I agree with Bernard: the function's defined on the whole real line and it clearly is an even one, thus it can not be injective. This is other way to attack this problem, imo. +1 $\endgroup$ – DonAntonio Feb 19 '17 at 23:42

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