Why is the unique mapping from $\emptyset$ to $Y$ inclusion? I am sorry, but I edited my question several times.
I am reading "Set Theory and General Topology" by Takeshi Saito(in Japanese).
The author wrote as follows:

When $X$ is a subset of $Y$, a mapping from $X$ to $Y$which maps $x \in X$ to $x \in Y$ is called inclusion. When $X$ is the empty set, the inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$.

$\forall x(x \in \emptyset \implies \text{every element } x \in \emptyset \text{ is mapped to } x \in Y \text{ by the mapping } \emptyset \to Y)$.
This is vacuously true.
But I think the following is also vacuously true:
$\forall x(x \in \emptyset \implies \text{every element } x \in \emptyset \text{ is mapped to } y \in Y \text{ such that } y \neq x \text{ by the mapping } \emptyset \to Y).$
So, the mapping $\emptyset \to Y$ is both inclusion and non-inclusion.
Then, the author could write as follows:

When $X$ is a subset of $Y$, a mapping from $X$ to $Y$which maps $x \in X$ to $x \in Y$ is called inclusion. When $X$ is the empty set, the non-inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$.

But the author wrote:

the inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$.

Why?
 A: Because $\emptyset$ is a subset of $Y$ and the unique application maps every element of the $\emptyset$ in itself. Clearly this requirement is void since the empty set has no element, so it is automatically satisfied.
I don't know if it's useful but a way you can define a function is this:
A function from $X$ to $Y$ is a subset $F$ of the cartesian product $X \times Y$ such that $$\forall x \in X \exists ! y \in Y | (x,y) \in F$$
(To be a bit more precise here I'm defining the graphic of a function, but if you fix domain and codomain, it's the same thing)
With this formalism you can define an inclusion mapping from $X$ to $Y$ as:
$$\Delta_X=\{(x,x)|x \in X\}$$
Where $ \Delta_X$ is called the diagonal of $X$. If $X=\emptyset$: clearly:
$$\Delta_{\emptyset}=\emptyset$$
And this defines the immersion of $\emptyset$ in $Y$
Regarding the edit, yes they are all vacuosly true. It's even true that $x \in \emptyset$ is mapped into a tomato or a cow. The fact that the second one is true doesn't imply that tue first one is not true.
I'll say you more, if you consider $f: \emptyset \to \mathbb{R}$ , this function is technically increasing and decreasing at the same time. The empty sets element satisfy any property(except existence) because every property becomes null.
On the second edit. The definition of inclusion is formally:
$\forall x \in X, f(x)=x$
Now clearly:
$$\forall x \in \emptyset, f(x)=x$$
So $f$ is an inclusion. Now you are saying tha since:
$$\forall x \in \emptyset, f(x)=tomato$$
Then $f$ is a non-inclusion. But this is technically wrong. The latter proposition isn't the negation of the former. The negation of the former is:
$$\exists x \in \emptyset | f(x)≠x$$
The "tomato" proposition doesn't imply this one if the set is void.
A: It holds through vacuous reasoning - loosely, vacuous reasoning is like "you can't prove me wrong."
Can you find $x,y \in \varnothing$ such that, for a mapping $f : \varnothing \to Y$, $f(y) = f(x)$ does not imply $y=x$? You can't. Or, phrased differently: $\forall x,y \in \varnothing$ $x \ne y \implies f(x) \ne f(y)$, because there are no violations to this (you can't find distinct $x,y \in \varnothing$ in the first place!).
An alternative justification: we know there exists an injection $f : A \to B$ whenever $|A| \le |B|$. Take $A = \varnothing$; then $|A|=0$ and there always exists some injection $f$ (since $|B| \ge 0$ for any set $B$).
A: Because it satisfies the definition.
$\emptyset$ is a subset of $Y$ and every element $x\in\emptyset$ is mapped to $x\in Y$.
It's kind of weird because $\emptyset$ has no elements, but precisely because it has no elements then every one of its elements satisfies that condition.
Check out the wikipedia article for vacuous truth.
A: The empty function being an inclusion is defined by the first statement being true, not by the second statement being false. Indeed, there are many functions where the second statement is false, but which are not inclusion functions.
The second statement describes a function without fixed point, and the empty function indeed has no fixed point.
The fact that both statements are (vacuously) true at the same time for the empty function, but not for any other function (with the empty set in the statement replaced by the corresponding domain), tells you that the empty function is the only inclusion function that has no fixed point. Which makes sense since all points in the domain of an inclusion function are fixed points, therefore the only way to have no fixed points in an inclusion function is to have no points in the domain, that is, to have an empty domain.
