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I musti proof the following:

" let $f: A \to B$, if $f$ is injection then $\forall a\in B \exists ! b\in A (f(b) = a)$"

proof: by contraddiction, therefore the negation of $\forall a\in B \exists ! b\in A (f(b) = a) $ is $\forall a \in B \exists c \in A (f(c) = a$ and $ c \neq b)$, but by hypothesis $f$ is bijection, injection and surjection, so $c \neq b$ then $f(c) \neq f(b)$, but $f(b) = a $ and $f(c) = a $ then $a \neq a$.. a absurd!!

Is correct?

Thanks in advance!

mle
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1 Answers1

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First of all: Why “$a ∈ B$” and “b ∈ A”? Why? And why “$\triangleq$”?

Secondly, your statement is wrong: It basically says that every injective function is bijective. But there are easy counter-examples:

  • Set $A = [0..∞)$, $B = ℝ$ and let $f$ be $[0..∞) → ℝ, x ↦ x$. Then $f$ is certainly injective but for $a = -1 ∈ B$ there is no $b ∈ A$ such that $f(b) = a$, or much simpler:
  • $\{1\} → \{1,2\}, \, 1 ↦ 1$ or even simpler:
  • the empty map $∅ → B$ for any nonempty set $B$.

And thirdly, the negation of $∀a ∈ B\, ∃! b ∈ A:\, (f(b) = a)$ is really: $$∃a ∈ B\, \big(∀b ∈ A:\, f(b) ≠ a \big) ∨ \big( ∃ b_1, b_2 ∈ A:\, (f(b_1) = a ∧ f(b_2) = a ∧ b_1 ≠ b_2)\big)$$ And in English: There is an $a ∈ B$ which is either not hit by $f$ or hit by two different $b_1$, $b_2$ in $A$.

Also, your argument is hard to follow and does not seem very logical. But my guess is that you mixed up some notions.

k.stm
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  • I edited it... but in this post: http://math.stackexchange.com/questions/228285/how-can-i-get-the-negation-of-exists-unique-existential-quantification the negation of exists unique existential is thinked intuitively.. – mle May 12 '13 at 08:42
  • Therefore in my hypothesis $f$ must to be bijective? – mle May 12 '13 at 08:44
  • @GarnakOlegovitc Well, I cannot know what you are supposed to prove, but it would at least make more sense. Do you see why that porperty of functions, viz to have a unique preimage, is in fact equivalent to being bijective? And do you see that your statement – as you formulated it – is indeed false? – k.stm May 12 '13 at 08:50
  • Now I understand! – mle May 12 '13 at 08:52
  • Well, … glad to help, I guess? – k.stm May 12 '13 at 08:53
  • of course.....!! – mle May 12 '13 at 08:54