injective

Injective (one-to-one) function

A function $f:A\to B$ is called injective if $\forall x_1,x_2\in A:f(x_1)=f(x_2)⟹x_1=x_2$
Or conversely: $\forall x_1,x_2\in A:x_1\ne x_2⟹f(x_1)\ne f(x_2)$

(an injective but not surjective function)

“uniqueness”

EXAMPLE

$f_1:\mathbb{R}\to {\mathbb{R}}_0^{+}:x\mapsto x^2$ is not injective, because $f_1(-1)=f_1(1)$ and it is not surjective, because $f_1(x)\ge 0\forall x\in \mathbb{R}$ (negative real numbers are not in the codomain).
$f_2:{\mathbb{R}}_0^{+}\to \mathbb{R},x\mapsto x^2$ is injective, but not surjective because we canot reach negative numbers.
$f_3:\mathbb{R}\to {\mathbb{R}}_0^{+},x\mapsto x^2$ is not injective but surjective ($x^2=z$ has a solution for all $z\in {\mathbb{R}}_0^{+}$).
$f_4:{\mathbb{R}}_0^{+}\to {\mathbb{R}}_0^{+},x\mapsto x^2$ is bijective.

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$f_1:\mathbb{N}\to \mathbb{N}:n\mapsto n+1$ is a bijection
$f_2:n\to \mathbb{N}:n\mapsto n-1$ is not a function, since $f(0)$ is not defined.
$g_1:\mathbb{Z}\to \mathbb{Z}:n\mapsto -n$ is a bijection
$g_2:\mathbb{Z}\to \mathbb{Z}:n\mapsto 1-n$ is a bijection
$h_1:{\mathbb{R}}^2\to {\mathbb{R}}^2:(x,y)\mapsto (y,-x)$ is a bijection (it’s a 90 degree rotation). Also, $x_1=x_2⟹y_1\ne y_2$ and vice versa.

$h_2:{\mathbb{R}}^2\to {\mathbb{R}}^2:(x,y)\mapsto (x-y,x+y)$ (hint: set up systems of equations) Injectivity: Assuming equal outputs for equal inputs:

\begin{align*}
(x_{1}, y_{1}), (x_{2}, y_{2}) \in \mathbb{R}^{2}: & h_{2}(x_{1}, y_{1}) = h_{2}(x_{2}, y_{2}) \
& \iff (x_{1} - y_{1}, x_{1} + y_{1}) = (x_{2} - y_{2}, x_{2} + y_{2}) \
& \iff \begin{cases}
x_{1} - y_{1} = x_{2} - y_{2} \
x_{1} + y_{1} = x_{2} + y_{2}
\end{cases} \
& \iff \begin{cases}
x_{1} = x_{2} \
y_{1} = y_{2}
\end{cases} \
& \iff (x_{1}, y_{1}) = (x_{2}, y_{2})
\end{align*}

Surjectivity: For every $(a,b)\in {\mathbb{R}}^2$, we can find a $(x,y)\in {\mathbb{R}}^2$ such that $h_2(x,y)=(a,b)$:

\begin{align*}
(a,b) \in \mathbb{R}^{2}: & h_{2}(x,y) = (a,b) \
& \iff (x-y, x+y) = (a,b) \
& \iff \begin{cases}
x - y = a \
x + y = b
\end{cases} \
& \iff \begin{cases}
x = \frac{a+b}{2} \
y = \frac{b-a}{2}
\end{cases} \
& \iff (x,y) = \left( \frac{a+b}{2}, \frac{b-a}{2} \right)
\end{align*}

→ $h_{2}$ is bijective.

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