I have a function $f :\mathbb R \times\mathbb R\to\mathbb R$ defined by $f(x,y)=x+y$ and I am struggling to determine if it is surjective or not. I already showed it is not injective. I know that if it is surjective I need to find some $z$ inside $\mathbb R$ such that $z=f(x,y)$. So then I have $z=x+y \Leftrightarrow z-x=y$. so Then $f(x,z-x)=x+(z-x)\Leftrightarrow f(x,z-x)=z$. I'm just not sure if this is correct or not.
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3$f(z, 0) = z$, so ... – Martin R Oct 18 '22 at 13:27
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The function $f$ is surjective iff for all $z\in \mathbb{R}$, there exists $(x,y)\in {\rm Dom}(f)$ such that $f(x,y)=z$. Then, $z=x+y$, notice that if $y=0$, then $z=x$ or if $x=0$, then $z=y$. So, for all $z$ there exists a point $(x,y)$ such that $f(x,y)=z$. Therefore, $f$ is surjective.
A. P.
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Notice that here surjective means "write a real number (here $z$) as a sum of two real numbers (here $x$ and $y$)". The statement sounds quite true and indeed it is. If you write $z = x + y$ where $z$ is the number you want to write as the sum of other two, then showing that $f$ is surjective means that you can find $(x,y)$ in the domain, so we are left on $how$ to find such $x$ and $y$. Quite simple: for any choice of $x$, the $y$ we are searching for is solution to the equation $z=x+y$, thus $y=z-x$. This means that the couple $(x,z-x)$ is a couple in the domain of $x$ such that $f(x,z-x)=z$ as we wanted.
Another way to see this function is clearly surjective is that the limit for $\|(x,y)\|\rightarrow+\infty$ of $f(x,y)$ diverges, and as $f$ is continous then it picks all real values from $-\infty$ to $+\infty$
Davide Trono
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A function is surjective iff its codomain and its range (or image) coincide.
The codomain of your function is $\mathbb{R}$ which concides with the range, hence the function is surjective.
Enrico M.
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What needs to be proven here is that the range actually is $\mathbb{R}$. – Paŭlo Ebermann Oct 18 '22 at 22:39