Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$ :

$$
f(x+y)=f(x)-f(y)
$$
By exchanging $x$ and $y$, we obtain that for all $x, y \in \mathbb{R}, f(y+x)=f(y)-f(x)$. Adding the initial equation, we get that $2 f(x+y)$ is always zero, so $f$ is identically zero (we can see this by setting $y=0$ for example).

Another way to find this result is to set $y=x$, which gives $f(2 x)=0$ for all $x$, with $2 x$ covering $\mathbb{R}$ as $x$ covers $\mathbb{R}$. Don't forget to check that the zero function is indeed a solution!

## Remark 5.

To guide one's research, it is often useful to try to find functions that are solutions, or particular values of the function using substitutions.

## - Injectivity/Surjectivity -

## Definition 6.

Let $f: A \rightarrow B$ be a function. We say that $f$ is injective if for every $b \in B$, the equation $f(x)=b$ has at most one solution in $A$. In other words, distinct elements of $A$ have distinct images:

$$
\forall a, a^{\prime} \in A, \quad a \neq a^{\prime} \Longrightarrow f(a) \neq f\left(a^{\prime}\right)
$$

## Example 7.

The identity function is injective, but the function $f: \left\lvert\, \begin{aligned} & \mathbb{R} \rightarrow \mathbb{R} \\ & x \mapsto x^{2}\end{aligned}\right.$ is not injective because $f(1)=$ $f(-1)=1$.

## Remark 8.

To prove that a function is injective, we show the contrapositive: if $f(a)=f\left(a^{\prime}\right)$, then $a=a^{\prime}$. In an exercise, if we show that the function $f$ is injective and we obtain an equality of the form $f($ thing $)=f($ widget $)$, we can therefore conclude that thing $=$ widget.