Is $f:\emptyset \to X$ injective My book uses 2 equivalent definitions of injectivity, the first being 
$$x\neq y \Rightarrow g(x)\neq g(y)$$
and the second being
$$g(x)=g(y) \Rightarrow x=y$$
Now as $f$ has $\emptyset$ as its domain I cannot make sense of  either of these definitions as i cannot put in a variable to actually test either of these. 
 A: Note that in your definition, you implicitly quantify your variables $x$ and $y$ over your domain. That is, your definition is:
$$
\forall x,y \in \emptyset , x \neq y \implies g(x) \neq g(y).
$$
Since there are no elements in the null set, this statement holds for every element in the null set (none), and so the function is indeed injective.
A: Yes, the function is injective. One way to see this is to make the definition more explicit:
A function $f:X\rightarrow Y$ is injective if 
$$\forall x \in X, \forall y \in X, (x \neq y \Rightarrow f(x) \neq f(y))$$
or equivalently (the contrapositive) if 
$$\forall x \in X, \forall y\in X, (f(x) = f(y) \Rightarrow x = y).$$
Whenever $S$ is an empty set, the statement $\forall x \in S, \ldots$ is always true—  vacuously true.  Such a statement says "Whenever you can find points in $S$ such that …", and because you can't find any points in an empty set $S$, the statement doesn't need to be checked for any points; it automatically holds.
The definition of injectivity is like this when the domain of $f$ is empty. It says "For any two points in the domain, …". The domain is empty so the statement doesn't need to be checked for any points; it automatically holds.
$f:\varnothing\rightarrow Y$ is injective.
