Let f be an injective function, that is:
$f : X \rightarrow Y$
$f(a) = f(b) \implies a = b$
Now, my question is, does the following need to hold in order for function to be injective:
$(\forall x \in X)(\exists y \in Y) (x,y) \in f$ (if we consider function to be a set of ordered pairs) (denote this statement by $(*)$)
1st case
If it does, then in order for function to be bijective, it needs also to be injective. For example in this case the function $f(x) = \frac 1 x$ is not injective, because it is not defined for x = 0 and $(*)$ does not hold; therefore it's also not bijective and inverse function does not exist. But $f^{-1}(x) = \frac 1 x$ .
2nd case
If statement $(*)$ is not required for function to be injective, then the definition of bijective function to be one-to-one correspondence does not hold, since there can be elements in domain that are not paired with any elements in range. So if we wanted to keep the definition of injective function without $(*)$, we would than have to redefine or rather extend bijective function not only as injective and surjective, but also satisfying $(*)$.
It seems to me to be intuitive paradox, but I'm sure I have made a mistake somewhere and I'd be greatful if someone explained it to me 😀
Best Answer
If for a given $x\in X$ there is no $y\in Y$ s.t. $(x,y) \in f$ then $x$ isn't in the domain of $f$. At that point saying $f:X\rightarrow Y$ is somewhat misleading. The domain of $f(x)=\frac{1}{x}$ is $\mathbb R\backslash \{0\}$.