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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

user9026
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  • Welcome to Mathematics Stack Exchange. What about the case where the function is not one-to-one, but not all elements are mapped to a single element? – J. W. Tanner Jun 07 '20 at 11:23
  • Sounds good to me! But I think you can give a shorter proof reasoning by contradiction and using the definition of function. – Álvaro G. Tenorio Jun 07 '20 at 11:24
  • Thanks for welcome. I get lot of my questions deleted in programming forums of StackOverflow. I hope math community is more tolerant. – user9026 Jun 07 '20 at 11:56

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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 $$|A|\geq|A'|=|f(A)|\ .$$

Christian Blatter
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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 $\;\stackrel{\text{by definition}}\implies |f(A)|\le |A|\;$

DonAntonio
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    $\phi$ is not well defined: for $b \in f(A)$ there may be several $a \in A$ such that $b=f(a)$. – mathcounterexamples.net Jun 07 '20 at 11:31
  • @mathcounterexamples.net I'd agree with you if I'd try to define for some general $;b\in f(A);$ , yet I choose to define for $;f(a)\in f(A);$ , which is tantamount, imo, to choose a definite element in the bundle $;b=f(a)\in f(A);$ . Anyway, in order to avoid confusion, I sall make some precisions. Thanks. – DonAntonio Jun 07 '20 at 14:23
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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 $$ |A| = \sum_{b\in B} N(b) = \sum_{b\in f(A)} N(b) \ge \sum_{b\in f(A)} 1 = |f(A)| $$

lhf
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  • Adapted from https://math.stackexchange.com/a/989059/589 – lhf Jun 07 '20 at 11:41
  • Does the proof assume that all elements of $A$ mapped to single element of $B$ ? Why $| f(A)| = 1$ ? – user9026 Jun 07 '20 at 11:54
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    @user9026, no it does not. The last line is "the sum of $1$ over all $b \in f(A)$ is equal to the size of $f(A)$" . – lhf Jun 07 '20 at 12:04