Surjective or not

[torpid vapor]

Is this surjective? I think this is what i'm supposed to do: Suppose y maps to f(x) then we have f(x)=y with y $\in\mathbb{R}$ and then solving for x we get $x=\sqrt[3]{2y-5}$ which to me seems surjective, as both negative and non-negative numbers map here and i think its pretty clear that it spans $\in \mathbb{R}$

[stable basin]

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[jolly pier]

Joshii

[torpid vapor]

am i missing steps

am i supposed to check $f(\sqrt[3]{2y-5}$

[jolly pier]

Joshii

[torpid vapor]

hm

in which case

$f(\sqrt[3]{2y-5})=\frac{1}{2}(2y-5+5})$

[jolly pier]

Joshii
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[torpid vapor]

which equals y

[cedar bridge]

assuming $f\colon\mathbb{R}\to\mathbb{R}$, then yes, for every $y\in\mathbb{R}$, $x:=\sqrt[3]{2y-5}\in\mathbb{R}$ and $f(x)=y$

[jolly pier]

Omegabet_

[cedar bridge]

that's the definition of surjections

[torpid vapor]

does checking if f(f(x)=y)=y ensure that its a surjection or noy?

[cedar bridge]

thats complete nonsense

you cant plug in "f(x)=y" into f

but like I literally just said

the definition of surjection is that, for $f:X\to Y$, $f(X)=Y$

[jolly pier]

Omegabet_

[cedar bridge]

$f(X)\subseteq Y$ is trivial, so the only thing to prove is $Y\subseteq f(X)$, ie for every $y\in Y$, there exists $x\in X$ such that $f(x)=y$

[jolly pier]

Omegabet_

[cedar bridge]

$f(X):={f(x)|x\in X}$ is the image of $f$ as typical.

[jolly pier]

Omegabet_

[torpid vapor]

thank you

+close