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I'm studying from a book and inside it I have this question:

Let $f:A\to B$ be a total function, which of the following states the $f$ is not injective:

A) For every $x,y \in A$ if $x=y$ then: $f(x) = f(y)$

B) There exist $f(x),f(y) \in B$ such that: $f(x) = f(y)$ and $x \ne y$

C) There exist $x,y \in A$ such that $f(x)=f(y)$ and $x \ne y$

First seeing this I almost instantly said that the correct answer is C, but in the book it says that the correct answer is B.

I kinda don't understand why B is true and C not, can anyone please tell me?

Thanks.

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    $\begingroup$ I agree with you and with @Justpassingby that C is the correct answer. B seems like a poorly-expressed version of C. What is the book you are studying from? $\endgroup$
    – mweiss
    Dec 9 '15 at 17:47
  • $\begingroup$ The correct answer is C, since B doesn't specify the domain of the variables. A function may nit be injective, but be injective when restricted to a smaller subset of the domain. $\endgroup$
    – Pedro Tamaroff
    Dec 9 '15 at 21:05
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I fully agree that C is the clearest expression of $f$ not being injective. B is ambiguous (it is not standard logic to use anything else than a variable after the existential quantor symbol $\exists$) but its only meaningful interpretation is equivalent to C.

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First seeing this I almost instantly said that the correct answer is C, but in the book it says that the correct answer is B.

I'm sorry to hear that. So much the worse for the book. I think anyone other than the book's author would say C. It's clearly equivalent to the negation of "$f$ is injective (on A)": $$\begin{align} \neg\, \forall x,y\in A\,(f(x)=f(y)\to x=y) &\iff \exists x,y\in A\,\neg\,(f(x)=f(y)\to x=y) \\ &\iff \exists x,y\in A\,(f(x)=f(y)\land x\ne y). \end{align}$$

B starts with the unusual phrase "there exist $f(x), f(y)$ in $B$...", shorthand for "there exist $u,v\in image(f)$ such that for some $x,y$ in $A$, $u = f(x)$ and $v=f(y)$ and ...". C is more straightforward, and doesn't even have to mention the codomain $B$.

A is true of every function, so it's not in the running.

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