Inverse function requirements 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 :D
 A: The definition of a function is
$$
\forall x \in X\  \exists! y \in Y:\ (x,y) \in f.
$$
The problem in your claim is that a map
$$
\mathbb{R} \rightarrow \mathbb{R}, x \mapsto x^{-1}
$$
does not exist. Thats why you have to specify $X$ and $Y$.
A: 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 $\mathbb R\backslash \{0\}$.
A: I think any bijection is defined to be both a surgection (onto) and an injection (1-to-1) at the same time.
As for your example, it matters to which sets you are mapping. For example, if we consider $f(x) = 1/x$ as a map $f:\mathbb{R}^+ \to \mathbb{R}^+$, all conditions for a bijection hold. Similarly for $\mathbb{R}-\{0\}$. However, for $\mathbb{R}$, the relation $f$ is not just a bijection, but it is not even a function, since functions require each input to be related to exactly one output, and $f(0) \not \in \mathbb{R}$.
One way to fix the problem is to consider $\mathfrak{R} = \mathbb{R} \cup \{\pm \infty\}$ and then $f$ is an injection (but not a surjection and not a bijection) on $\mathfrak{R}$. Finally, if you want a bijection, you may want to consider $f$ on $\mathfrak{R} = \mathbb{R} \cup \{\infty\}$ (or on $\mathbb{R}^+$ or $\mathbb{R}^-$ as we have already pointed out).
