Injective or surjective function $f(x,y)$ I have this exercise:
Show if the next function is injective, surjective or bijective.
$$f \colon \mathbb{R}^{2} \rightarrow \mathbb{R}, f(x,y)= xy. $$
I know the injective function is 1 by 1, i,e. $f(a)=f(b) \implies a=b \quad$ and surjective is a function that maps an element $x$ to every element $y$; that is, for every $y$, there is an $x$ such that $f(x) = y$.
My doubt in the part of $\mathbb{R}^{2} \rightarrow \mathbb{R}$. I don't know how interpret that. Could you help me with any hint, please?
 A: It is definetly not injective, since $f(1,2)=f(2,1)=2$. More generally, you cannot have an injective map from space $A$ to space $B$ if the dimension of $A$ is higher than the dimension of $B$. Intuitively: you have many more elements in $\mathbb{R}^{2}$ than in $\mathbb{R}$.
However, $f$ is surjective: just fix $y=1$ and then for any $z$ the pair $(z,1)$ does the job. Finally, since it is not injective it cannot be bijective.
A: The function is surjective, but not injective. Therefore, it is not bijective.

*

*For every $x\in \mathbb R$, there exists $(x,1)\in \mathbb R^2$ such that $$f(x,1) = x\cdot 1 = x$$
so, $f$ is surjective.


*Take $(x_1,y_1) = (1,6)$ and $(x_2,y_2) = (2,3)$. Then, $(x_1,y_1)\ne (x_2,y_2)$ but $$f(x_1,y_1) = x_1y_1 = 6 = x_2y_2 = f(x_2,y_2)$$
so, $f$ is not injective.
A: The wording of the questions suggests you're confused by the $f:\mathbb{R^2}\rightarrow\mathbb{R}$ part. For any set $X$, the notation $X^2$ represents the Cartesian product of $X$ with itself, or in other words the set of all pairs of elements of $X$. In set notation:
$$X^2 = \{(x, y) : x \in X \land y \in X\}$$
So $\mathbb{R}^2$ just means "the set of all pairs of real numbers $(x, y)$". Thus, $\mathbb{R}^2 \rightarrow \mathbb{R}$ denotes "this function takes two real numbers as input, and produces a single real number as output". Thus the "typical" input of $f$ is a pair of real numbers $(x, y)$, and its "typical" output is a single real number, say $z$.
To check whether $f$ is injective, you need to test whether every output of $f$ is produced by a unique input - in other words, if $f(x_1, y_1) = f(x_2, y_2)$, is it possible to have $(x_1, y_1) \neq (x_2, y_2)$?
Likewise, to check whether $f$ is surjective, you need to test whether every possible output is produced by some input. In other words, given $z \in \mathbb{R}$, can you always find $(x, y) \in \mathbb{R}^2$ such that $f(x, y) = z$?
A: Proceed the same, with your same reasoning. In this case $ f $ is injective if and only if from $ f (x, y) = f (a, b) $ it follows that $ x = a $ and $ y = b $. Then analyze whether or not it is injective. For surjectivity, you would only have to see if each real $ z $ is of the form $ xy $, which certainly does happen ($f (z, 1) = z$).
