surjective function

Depends what the co domain is. If the function is say from $\mathbb{R}\rightarrow \mathbb{C}$ then it is not surjective. A function is surjective if the range or image is equal to the co domain.

well it goes from R-->R

Well in that case yes it is surjective.

so far i'm looking for a function which is surjective but not surjective when squared

constants don't work, they're neither injective or surjective

fraction functions don't work like 1/x

minus function are pretty much the same as positive ones they just flip...

$\sqrt x$???? square it, it's bijective but then i(root x)t fails the vertical line test .__.


i'm thinking still don't tell me an answer tho ^_^

Remember that functions don't have to be defined for all values of $\mathbb{R}$, functions can be defined on any subset of R indeed a finite subset as well.

$\sqrt x$???? square it, it's bijective but then i(root x)t fails the vertical line test .__.

i'm thinking still don't tell me an answer tho ^_^

but they do tho? like i can't choose 1/x right? cos it doesn't work in the realm of R-->R

Also $\sqrt x$ doesn't fail the vertical line test.

You could have a function f such that f(x)=1/x for all x in R except 0, and then not have f defined at 0. Or you could define f(0) to be whatever number you want -it's still a function.
Consider a function $f:\{-1,1\}\rightarrow \{-1,1\}, x\mapsto x$, this is a perfectly well defined function and is surjective (bijective in fact).

agreed but it doesn't fit in with the statement at hand which is

"if $f:\mathbb R \Rightarrow \mathbb R$ is surjective, the so is $g(x) :=[f(x)]^2$"

what i need to find if this is true or false, i think it's false so i have to find 1 example only of which this doesn't work ..

What happens if f(x) is negative?

agreed but it doesn't fit in with the statement at hand which is

"if $f:\mathbb R \Rightarrow \mathbb R$ is surjective, the so is $g(x) :=[f(x)]^2$"

what i need to find if this is true or false, i think it's false so i have to find 1 example only of which this doesn't work ..

I didn't know the function had to be from R to R.
Why not take the most simple example I can think of, f(x)=x, in that case g(x)=x^2 which is not surjective on R.
Also $x\mapsto \sqrt x$ is not a valid function from R to R.

still in the set of reals but it's pretty much just flipped

then again root x seems lke neither now tho cus it feels like if i pick any x value in the set of reals then for root x there's isn't always a y to correspond simply 1 to 1 or just a value(any will do).

I'm thinking modulus now

that's the one i used for the previous question which had injective instead of surjective

i'm sure that means i gotta find a function of which surjective is a subset of it and a function which is injective only i'm thinking modulus now but ye root x don't work, it's not a function in R->R

x is bijective from R->R just pick any value of x and you pick the same value for y
now x² isn't injective bcus you have -1 and 1 which give out 1 the same value but 2 different inputs so that's just surjective

that's not wat i need, i was thinking x^3 and then x^6 but that's the same as x and x^2

I'm not sure you are getting what surjective means it means that the image covers entirely the set that you choose for the codomain. Therefore, in your problem you need to choose a codomain so that this equals the image of f(x) so that it is injective then you need to find real numbers like a lets say such that f(x)=a but there exists no x such that that f(x)^2=a and then since the image of f(x) equalled the codomain and so a is in the codomain then f(x)^2 doesn't equal the codomain.

What is your question exactly - I'm not sure what you are trying to do.

The question is, is

"if $f:\mathbb R \Rightarrow \mathbb R$ is surjective, the so is $g(x) :=[f(x)]^2$"

true or false?