# The precise definition of injective

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.

• 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? Dec 9 '15 at 17:47
• 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. Dec 9 '15 at 21:05

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.
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 $B$.