assume there is a function like $f:A→B$ which is injective, why it means $\left|A\right|\le\left|B\right|$ or in another way why a function like $g:B→A$ stands for $\left|B\right|\le\left|A\right|$
my problem has been clearly explained here: assume a set like $A=[1,2,3,4]$ and $B=[1,2,3]$ and for $f:A→B$ we have $f=[(1,1)(2,2),(3,3)]$ and $f$ is still injective but $\left|A\right|=4$ and $\left|B\right|=3$ and in this case $\left|B\right|\le\left|A\right|$...
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A function $f:A\rightarrow B$ is a relation $f\subseteq A\times B$ such that for each $a\in A$ there is exactly one $f(a)\in B$, i.e., $f=\{(a,f(a))\mid a\in A\}$.
Let $A,B$ be finite sets. Then the definition of $f$ shows that $|A|\leq |B|$.
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In old books you can find a difference between function and aplication.
- A function $f\colon A\to B$ has as domain a subset of $A$.
- An application $f\colon A \to B$ has as domain all $A$
for people, in general , those means the same.
If you want to speak strictly, the intuition behind injectivity is:
You can find a "copy" of the domain of $f\colon A \to B$ in the set $B$
In your example $f$ is not a vell defined function since $4$ is out of it domain, so it can not be injective since is not a function.
But you can make it a function as soon as you say $f(4)$ is something in $\{1,2,3\}$. But then it is not injective.
In general if $|B|<|A|$ then two different $a,b$ in $A$ must have the same picture, so $f$ can not be injective.