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I have a problem understanding the following proof that claims a surjection. The text is translated from a german university textbook by Luise Unger (pardon any translation errors by me, please).

Given $$f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$$ $$f((x,y)) = x+y, \text{for all} (x,y) \in \mathbb{R} \times \mathbb{R} $$

then $f$ is surjective, because if $z \in \mathbb{R}$, then $(0,z) \in \mathbb{R}\times\mathbb{R}$, and it is $f((0,z)) = 0+z = z$. Therefore, every $z\in\mathbb{R}$ has a fiber under $f$.

I understand the principle of surjective functions (that every value in the codomain is "reachable" through $f$). However, I don't understand how the truth of the special case $f((0,z))$ explains the truth of the (whole) surjection; or is that proof incorrect or incomplete?

phresnel
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  • Note that for an arbitrary value of $z$ we must be able to exhibit some pair $(x,y)$ such that $f(x,y)=z$. It does not matter if we pick a very special pair, provided we can do so for every value of $z$. – Hagen von Eitzen Oct 21 '13 at 22:14
  • Indeed, there are other ways to get the value of $z$, this is just the simplest. Actually, that's how you can see $f$ is not injective. – Tyler Oct 21 '13 at 22:15
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    As to your profile, the quote is: "Give a man a fire, he's warm for one night. Set fire to him, he's warm for the rest of his life." – Will Jagy Oct 21 '13 at 22:19
  • @Will Jagy: Even though google proposes both variants? What is with: http://en.wiktionary.org/wiki/give_a_man_a_fish_and_you_feed_him_for_a_day._Teach_a_man_to_fish_and_you_feed_him_for_a_lifetime – phresnel Oct 22 '13 at 06:02
  • @Tyler: Yes, she uses the exact same example to prove non-injection. – phresnel Oct 22 '13 at 06:03
  • phresnel, my version is in one of the Discworld books by Terry Pratchett, called Jingo. http://en.wikipedia.org/wiki/Jingo_%28novel%29 and http://www.goodreads.com/work/quotes/1128623-jingo – Will Jagy Oct 22 '13 at 06:09
  • @WillJagy: Ahh! Need to read more of him. Actually, I've just read Wintersmith, and hardly any book ever made me laugh in a full train. But that one did. – phresnel Oct 22 '13 at 06:14

1 Answers1

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To show that a function is surjective you need to show that given $y$ in the codomain, there is some $x$ in the domain such that $f(x)=y$.

In this case, $x$ is an ordered pair $(u,v)$. So we want to show that for every real number $y$ there is some ordered pair $(u,v)$ such that $u+v=y$. If we can exhibit such ordered pair for $y$, for every $y$, then the function is indeed surjective. And as the proof says, we can take the pair $(0,y)$.

This is not a particular case, it's the general case, because the generality here is for $y$, rather than for the ordered pairs.

Asaf Karagila
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    Asaf, you're a madman. Also, each fiber $f^{-1}(a)$ is the line $x+y = a.$ We need display only one point in the fiber to prove existence (non-emptiness) of the fiber. – Will Jagy Oct 21 '13 at 22:14
  • I am. That's true. Why this time, though? – Asaf Karagila Oct 21 '13 at 22:14
  • I'll think of something. Give me a few minutes. – Will Jagy Oct 21 '13 at 22:15
  • hmmm, cannot remark on your clothing as have no information... – Will Jagy Oct 21 '13 at 22:17
  • All black t-shirts (short and long sleeves); all but two are plain black; the other two have Guinness writing on them, one in a small font, the other as part of a pub banner. Two pairs of long black training pants; two pairs of short black (now dark gray) pants; two pairs of blue jeans. One "student jacket", black. Black and very long Guinness scarf for the extra cold days. – Asaf Karagila Oct 21 '13 at 22:19
  • Thank you, Asaf. Those do not really say "madman" to me either. I may have spoken too quickly. Do you go out in the midday sun? There is this thing about Englishmen... – Will Jagy Oct 21 '13 at 22:21
  • I love the sun. I am a desert child. Grew up in the scorching heat of south Israel. At the same time... I am also a night owl. Weird, but not yet madman quality. C'mon, dig deep! – Asaf Karagila Oct 21 '13 at 22:25
  • How about substance abuse? Heroin, peyote, anything along those lines? – Will Jagy Oct 21 '13 at 22:41
  • Got it. Ever want to murder your father and sleep with your mother? That's a good one. – Will Jagy Oct 21 '13 at 22:42
  • Penchant for good scotch, unusual capacity for drinking (although I hardly ever exploit it when I'm in Israel); other than that... not much. :-) – Asaf Karagila Oct 21 '13 at 22:42
  • I'm glad I took this course pass/fail, this is tougher than it looks. – Will Jagy Oct 21 '13 at 22:43
  • Seriously? Are you in freshman psychology here? Oedipus? – Asaf Karagila Oct 21 '13 at 22:43
  • I did take part of a psychology course, once. A math professor suggested a teaching certificate was a good adjunct to a bachelors. I'm not sure he realized the load of non-math stuff involved. I once went to a shopping mall and took notes on the groups of kids; they looked at me funny. I believe i managed to drop the course after that. – Will Jagy Oct 21 '13 at 22:46
  • Interesting conversation, Asaf, Will :) So, I think what struggled me was that the surjection says something about the domain/codomain, and not only the function. This is why she wrote "because of $z\in\mathbb{R}$", which is exactly the codomain. And because using the pair $(0,z)$, you can "cover" the whole codomain, the function is surjective. You now have the chance of making me happy by just saying "yes" :) – phresnel Oct 22 '13 at 05:57