# Thread: Injective / surjective function

1. ## Injective / surjective function

a. let f:R^2-->R^2 be given by f(x,y)=(x,xy) . If the mapping injective? is it surjective?

b. Represent F(a) by a picture when A - {(x,y) ∈ R^2 | x ∈ [1,2], y ∈ [0,2]}

2. Hint for part (a).

What can you say about $f(0,0)$ and $f(0,y)$ where $y\in\mathbb R,\,y\ne0\,?$ Also, can you find some $(a,b)\in\mathbb R^2$ such that $f(a,b)=(0,y)\quad(y\ne0)\,?$

3. Originally Posted by champrock
a. let f:R^2-->R^2 be given by f(x,y)=(x,xy) . If the mapping injective? is it surjective?
i) $f(0,1)=?~\&~f(0,2)=?$

ii) If $f(a,b)=(0,2)\text{ then }a=?~\&~b=?$

4. Originally Posted by Plato
i) $f(0,1)=?~\&~f(0,2)=?$

ii) If $f(a,b)=(0,2)\text{ then }a=?~\&~b=?$
Please let me know if this is correct:
1. the function is not Injective since f(0,1) = (0,0) and f(0,2) = (0,0). So, there exist differnt values in the domain which have the same image.

2. The function is surjective since if f(a,b) = (0,2) then it is always possible to find atleast one value of a & b which give that image?

5. Originally Posted by champrock
Please let me know if this is correct:
1. the function is not Injective since f(0,1) = (0,0) and f(0,2) = (0,0). So, there exist differnt values in the domain which have the same image.
Yes, that is correct and was Plato's and TheAbstractionist's point.

2. The function is surjective since if f(a,b) = (0,2) then it is always possible to find atleast one value of a & b which give that image?
NO. Showing that there exist a, b such that f(a,b)= (0,2) does not show it is surjective. You would need to show that for any numbers x, y, there exist a, b such that f(a,b)= (x, y).

And you did not show that there must be a, b such that f(a,b)= (0,2), you simply asserted it! If f(a,b)= (0, 2), HOW would you find a and b? Exactly what must they be?

6. And you did not show that there must be a, b such that f(a,b)= (0,2), you simply asserted it! If f(a,b)= (0, 2), HOW would you find a and b? Exactly what must they be?
OK ya, got it. There cannot be any values of a&b which give the image as (0,2). [Since if x=0 then x*y=0 always). So the fuction is not surjective. There does not exist any value in the domain which is giving this particular value in the image.