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I need to prove that $f$ is onto, where $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}; f(x,y)=(x+y,y)$.

I do understand the "principle" behind the meaning of onto functions.

Here is a proof for this problem:

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But I don't understand the conclusion. Why does proving that $x=x'-y'$ conclude the proof? Is it just because we showed that such an $x$ exists? Or does it actually matter the value in particular of $x'-y'$? What would it look like if the function $f$ were NOT onto (I am having trouble picturing a scenario when such $x$ does not exist)?

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    It might help to write down, with quantifiers and as formally as you can, what it means to say that $f$ is onto. Can you do this? – 2017-02-26
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    You are given an arbitrary tuple $(x',y')$ can you find $(x,y) \in \mathbb{R}^2$ to make $f$ onto? Yeah you did, $x = x' - y'$ and $y = y'$. – 2017-02-26
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    @Git Gud, yeah...I have proven functions are onto before...a long time ago. I think that since it's a tuple here, the format is throwing me off (and/or I only vaguely remember the process correctly). But, the more formal definition is that for each element $y$ in the codomain, prove that there exists an element $x$ in the domain such that $f(x)=y$. – 2017-02-26
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    @Nameless, so you are saying that it's just the fact that there exists a tuple that maps to $(x+y,y)$? I guess it makes sense. I don't know why this problem is uncomfortable for me. haha Maybe I am not sure why we just picked some arbitrary point that is equal to $(x+y,y)$. Is that what we regularly do when proving functions are onto? – 2017-02-26
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    @PBJ That's right. Changing the symbols for better readability: for every $v$ in the codomain, find $u$ such that $f(u)=v$. Here the domain and the codomain are $\mathbb R^2$. So for every $v$ in $\mathbb R^2$ you want to find $u$ in $\mathbb R^2$ such that $f(u)=v$. Since $v\in \mathbb R^2$, there exist $x', y'$ in $\mathbb R$ such that $v=(x', y')$. You wanna find $u$ in $\mathbb R^2$ s.t. $f(u)=(x', y')$. Since $u$ is supposed to be in $\mathbb R^2$, this is equivalent to finding real numbers $x$ and $y$ such that $f(x,y)=(x',y')$ (and you want to pick $u=(x,y)$). The rest is arithmetic. – 2017-02-26
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    Ok, I think I get it now. I wasn't thinking of the proof's conclusion in terms of ordered pairs, like @Nameless described it. I didn't realize the conclusion was actually to point out that the pair $(x'-y',y')$ maps the function. I was just thinking that since $x=x'-y'$ the direct calculation itself is what proved everything. – 2017-02-26
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    And, @Git Gud your explanation helped a lot. Thanks to both! – 2017-02-26
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    "so you are saying that..." No, "it's just the fact that", for every $(x',y')$, there exists a couple $(x,y)$ that maps to $(x',y')$. – 2017-02-26
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    It is a `linear map` from a vector space of finite dimension (2) to itself, and it is injective, hence it is surjective. – 2017-02-26
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    Suppose $f(x,y) = (x^2, y^2)$ so you have to find $(x',y')$ so that $x'^2 = x$ and $y'^2 = y$ for *ANY* $(x,y)$. Will that *always* be possible? – 2018-02-10

2 Answers 2

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What you have to understand is the following. Given $(x+y,y)\in\mathbb{R}^2$ does there always exist $(x,y)\in\mathbb{R}^2$ such that $T(x,y)=(x+y,y)$ ? This is what precisely @Nameless's 1st comment tells you.

Here is another way to prove onto property in case you are interested.

Let $T:\mathbb{R}^2\to \mathbb{R}^2$ be given as $T(x,y)=(x+y,y)$. So $T$ is a linear transformation. Let us first show that $T$ is one to one. This means $Tx=0\implies x=0$. Now $(x+y,y)=0\implies y=0$ and $x+y=0$ whence we get $x=0$. So $(x,y)=(0,0)$. Thus $T$ is one to one. It is a standard result that a linear transformation is one to one iff it is onto.

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To prove that $f$ is onto you must pick an arbitrary $(x, y) \in \mathbb R^2$.

If you can find an $(x',y')$ so that $f(x',y') = (x,y)$ then that proves $f$ onto, because if we can do it for $(x,y)$ we can do it for any other point because there was nothing special about $(x,y)$.

If we let $y' = y$ and $x' = x - y$then

$f(x',y') = (x'+ y', y') = ((x-y)+y, y) = (x,y)$.

We can do that for any point, ... say $(a,b)$ ... and we can always find $f(a-b, b) = (a,b)$ so we can map to any point.

Perhaps it would help to show how a function might NOT be onto.

Let $f(x,y) = (x^2 + y , y)$

We pick an $(x,y)$ arbitririly. And we want to show that there is always some $(x',y')$ such that $f(x',y') = (x,y)$. We we FAIL. Because that isn't true.

So we must solve for $f(x',y') = (x'^2 + y', y') = (x,y)$

So we must have that $y' = y$ and that $x'^2 + y' = x$.

But that means $x'^2 + y = x$ so that $x'^2 = x - y$. But what if $x - y < 0$. If that happened there's no such $x'$ that will make that true.

So if $y>x$ this is impossible that there is any $(x',y')$ so that $f(x',y') = (x,y)$. For example if $f(x,y) = (x^2 + y, y) = (0,1)$ is impossible. So $f$ is NOT unto.