Given $2$ finite sets exists, show that the following statement is true. I'm trying to get an early start on my discrete structures subject next semester (trying to get a head start :D) and trying to understand key parts of it. I was doing some online research about combinatorics when I found this interesting question. I figured this has something to do with set theory but I'm not sure which sub-topic this falls under. 

You're given a positive natural number, $a \in \mathbb N^+$, two finite sets $X,Y$ such that $|X| > a|Y|$, and $f: X \to Y$. Show that there exists a $y \in Y$ such that $|f^{-1}(y)| ≥ a+1$.

Before actually answering this question, I would be grateful for any tips on what I should search to get a better understanding of how to solve this kind of a question and how I should go about answering such a question.
PS - This is my first time using this forum so any suggestions of how to better make a question or what kind of further information I should put in the question, would be great! 
 A: There are some tips : 


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*It looks natural to reason by contraposition : i.e to suppose $|f^{-1}(\{y\})| \leq a$ and to show $|X| \leq a|Y|$ (there is no more $a+1$).

*Some intuitive idea : the $|f^{-1}(\{y\})|$ are called fibers of $f$. The application $f$ from $X$ to $Y$ is in a certain way giving some structure on the set $X$. Mentally, you can put the different $y$ horizontally, and the elements of a same fiber $f^{-1}(\{y\})$ vertically over $y$. f can be thought as a projection on the horizontal axis.

*It is linked to the previous point : the sets $f^{-1}(\{y\})$, for $y$ in $Y$, constitutes a partition of the set A, i.e the sets $f^{-1}(\{y\})$ are disjoint and their union is A. This implies something for the cardinal of the sum... 
A: The sets are finite so....
Think it out... Let $|Y| = n$ and $Y = \{y_1, ......, y_n\}$ and 
ANd each $f^{-1}(y_i) \subset X$ must be finite so let $|f^{-1}(y_i)| = K_i$
Now $f: X\to Y$ so $f^{-1}(Y) = X$ and $f^{-1}(Y) = \cup_{i=1....n}f^{-1}(y_i)$.
Now the basic inclusion exclusion property of finite sets says that $|A \cup B| = |A| + |B| - |A\cap B|$ so $|X| = |f^{-1}(Y)| = |\cup_{i=1....n}f^{-1}(y_i)| \le \sum_{i=1}^n |K_i|$.
Now $|X| > a|Y|= an$.
... Pigeon hole.... if each $|K_i| \le a$ then $\sum_{i=1}^n|K_i| \le \sum_{i=1}^n a = an < |X| \le  \sum_{i=1}^n |K_i|$.  A contradiction.
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In plainer english:
Suppose $Y$ has $n$ elements and $Y = \{y_1, ...., y_2\}$.  And $X$ has more than $an$ elements.
$K_i= f^{-1}(\{y_i\}$ is the set of all elements of $Y$ that get mapped into $y_i$.  The are $n$ of these sets:  $K_1 = f^{-1}(\{y_1\}), K_2=f^{-1}(\{y_2\}), ..... etc.$
If each of these sets have at most $a$ elements then in total these sets have at the very most $an$ elements.
But every element of $X$ gets mapped to some element of $Y$ (that's what a "function" means) so every element of $X$ is in one of these sets.  So at the absolute very most $X$ has $an$ elements.
Which contradicts our assumption that $X$ had more than $an$ elements.
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Maybe to get an intuition we can do a physical example:
Let $Y = \{1,2,3,4\}$ so $|Y| = 4$ and $a = 3$ and $X= \{1,2,3,4,......, 13\}$ and $|X| = 13 > 3*4$.
Let say for $x \in \{1,2,3,4,.......,13\}=X$ then $f(x) \in \{1,2,3,4\} = Y$.
Consider $\{f(1), f(2), f(3),....., f(13)\}$.  As there  are only $4$ possible values these can be there must be some repeats.  How many repeats?  Well let's suppose $f(x) = 1$ has $3$ repeats.  There are there $x_1,x_2,x_3 \in X$ so that $f(x_1)=f(x_2) =f(x_3) = 1$.  And $f^{-1}(\{1\} ) = \{x_1, x_2, x_3\}$.
Okay, that's fine. Now how many repeats does $f(x) =2$ have?  Let's say it has $3$.  So there are $x_4,x_5,x_6$ so that $f(x_4) = f(x_5)=f(x_6) = 2$.
And how many repeats does $f(x) = 3$ have? and $f(x) =4$?  Assume those have $3$ each too.
So how many $x \in X$ have we accounted for $x_1,x_2, x_3$ all map to $1$.  And $x_4,x_5, x_6$ all map to $2$.  And $x_7,x_8, x_9$ all map to $3$.  And $x_{10}, x_{11}$ and $x_{12}$ map to $4$.
But... there are $13$ $x \in X$.  Which one did we not account for?  If $13$ elements are mapped into $4$ options there must be at least one option with more than $3$ elements mapped to it.  

Another way of looking at it is if $|X|> na$ elements are mapped to $|Y|=n$ options, each option has on average $\frac {|X|}{|Y|} > \frac {na}{n} = a$ elements mapped to it.  Now it's not possible for every option to have fewer than average mapped to it.  So only must have at least the average mapped to it.  And the average is at least $a + 1$.
