If $f:A \to B$ then prove that $|A| \geqslant |f(A)| $ So, $A$ and $B$ are non empty finite sets and there is a function $f:A \to B$ and I need to prove that $|A| \geqslant |f(A)| $.  So, $|A|$ is the cardinality of set $A$ and $ |f(A)| $ is the cardinality of the co-domain of the function. So, here is my thinking on this. Since $f$ is a function, each element in $A$ is mapped to some element in $B$. In the worst case, all elements of $A$ are mapped to a single element of $B$. And since $A$ is non-empty, there is at least one element in $A$. So, if there is only one element in $A$, then we have $|A| = |f(A)| $. Another case is when $|A| > 1$ and $ |f(A)| = 1$. Here we would have $|A| > |f(A)| $. And another case is when the function is one to one. So, every element of $A$ must be mapped to a different element of $B$. So, we must have $|A| = |f(A)|$. So, in any case, we have $|A| \geqslant |f(A)| $. Now, would this be considered a valid proof ? I don't know if this proof is without any loopholes. 
Thanks
 A: For each $y\in f(A)$ choose an $x\in A$ with $f(x)=y$. In this way  $|f(A)|$ different points $x\in A$ are chosen, forming a subset $A'\subset A$. We then have
$$|A|\geq|A'|=|f(A)|\ .$$
A: Directly is, imo, much simpler and short: 
Define $\;\phi: f(A)\to A\;$ as follows: for any element $\;x\in f(A)\;$ choose one single element $\;a_x\in A\;$ s.t. $\;f(a_x)=x\;$. Then define $\;\phi(x):=a_x\;$ (this may require $\;AC\;$ in case $\;A\;$ is infinite), 
$$\phi(x)=\phi(y)\implies a_x=a_y\stackrel{\text{because $f$ is a funct.}}\implies f(a_x)=f(a_y)\implies x=y\;$$ 
and thus $\;\phi\;$ is injective $\;\stackrel{\text{by definition}}\implies |f(A)|\le |A|\;$
A: For each $b\in B$, let $N(b)$ be the number of elements in $A$ that are mapped to $b$ by $f$;
$$
N(b) = | \{ a \in A : f(a)=b \} |
$$
Then $\sum_{b\in B} N(b) = |A|$ because every element of $A$ is mapped to some element of $B$. (This is the statement that the fibers of $f$ partition $A$.)
Therefore
$$
|A| = \sum_{b\in B} N(b) =  \sum_{b\in f(A)} N(b) \ge \sum_{b\in f(A)} 1 = |f(A)|
$$
