#Set theory

1 messages · Page 1 of 1 (latest)

leaden steeple
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I really need help with a math question:
We have f1 R^2 -> R and f2 R^2 -> R maps and define f: R^2 -> R^2 with f(x, y) = (f1(x, y), f2(x, y)).
a) Suppose that f is surjective, prove that f1 and f2 are surjective.
b) Suppose that f1 and f2 are surjective, Is f then also surjective? Prove this or provide a counterexample.

shrewd valve
leaden steeple
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In mathematics, a surjective function (also known as surjection, or onto function) is a function f that maps an element x to every element y; that is, for every y, there is an x such that f(x) = y. In other words, every element of the function's codomain is the image of at least one element of its domain.

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copy from wikipedia

shrewd valve
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Right.

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So if your two functions f1 and f2 are surjective, what does that mean?

leaden steeple
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intuitively it should imply f is also surjective, with examples its very obvious, but i just cant put it formally on paper

shrewd valve
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Wait, let's back up. Did you need help with a?

leaden steeple
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I cant figure both of them out

shrewd valve
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Okay, well, let's take them one at a time.

leaden steeple
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Sure first A then i guess

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I would start like: if for any a, b there exists x, y such that f(x, y) = (a, b). I would have to show that for any a, there exists x, y such that f1(x,y) = a and there exists x, y such that f2(x,y) = b

shrewd valve
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So what does it mean if f is surjective? Like, by definition?

shrewd valve
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Or rather, how have you already showed that but not noticed yet?

leaden steeple
shrewd valve
leaden steeple
shrewd valve
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Right, so if, for all real numbers a and b, there exists some real numbers x and y such that f(x, y) = (a, b), that means that f1(x, y) = a and f2(x, y) = b.

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Because f(x, y) = (f1(x, y), f2(x, y)).

shrewd valve
leaden steeple
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Yes

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But this isn’t a proof right

shrewd valve
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I mean.

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It pretty much is.