I asked myself and also proved both theoritically and intuitively that we have the following implication. Please prove me right or wrong.
$$(f:E\rightarrow F\quad is\quad a\quad bijection)\Rightarrow (Card(E)\le Card(F))$$
And here follows my proof:
Let $E$ and $F$ be two arbitrary sets.
Assume that $f:E\rightarrow F$ is a bijection. Then, we have;
$$\forall x\in E,\quad \exists !y\in F\quad |\quad f(x)=y$$
Lets choose any $x\in E$, so that we have $f(x)\in F$.
We have here that $(x,f(x))\in (E\times F)$ is a unique couple.
Thus,
$$Card(F)=Card\left\{ \forall x\in E, (x,f(x)) \right\} $$
We will have to study now the two case between $Card(E)$ and $Card\left\{ \forall x\in E, (x,f(x)) \right\}$:
1st case:
Have we any $x\in E$ such that $f(x)=\emptyset$? But, obviously, this will result into a contradiction with our assumption for bijective definition of the function $f$, because of the unicity & existence are part of predicate.
2nd case:
Have we any $y\in F$ such that $f^{ -1 }\left( y \right)=\emptyset$? We can though easily observe that existence of such $y$ is obvious, because we would conclude then that $f$ is injective -which is included within assumption.
So, there may exist at least one such $y\in F$ which doesn't have reciprocal image in $E$
Consequently,
We deduce that;
$$Card(F)\ge Card(E)$$ $$\Box $$
PS: I've forced myself for a forward implicative reasonning, though it would be easier by contradiction perhaps.
I thank you a lot for commenting my question. Indeed, the error on my behalf was to consider wrong assumptions regarding definitions I used.
I'm writing this prompt reply to make it clear that, resuming commentors' remarks:
It's a function which is defined as: $$\forall x\in E,\quad \exists !y\in F\quad |\quad f(x)=y$$
But for bijection, we have: $$\forall y\in F,\quad \exists !x\in E\quad |\quad f(x)=y$$
And, for the relation between bijectivity and cardinilities are exactly:
Thank you very much again.