Consider the set $[n]=\{1,2,\ldots,n\}$ and $\mathcal C^k=\{A\subset [n],\ |A|=k\}$. Given two integer numbers $k<l$, is there a one-to-one function $\varphi\colon \mathcal C^l\to \mathcal C^k$ such that $\varphi(A)\subset A$?
Of course, since $|\mathcal C^i|=\binom ni$ we have a necessary condition $\binom nk\geq \binom nl$. Is it also a sufficient condition?
This statement is so clear and the problem so well formulated that I believe the answer is well known.
I came accross this problem when solving some old high-school math-olympiad problem. There was some combinatorial issue and I managed to solve it using inclusion–exclusion principle, but in order to do this, I had to know that I can assign to any $4$-element subset of $\{1,2,3,4,5,6\}$ its $3$-element subset and this assignment is an injection. I did it by hand (there are only 15 subsets having 4 elements so it's not a big task). I hope there is more elegant method and I'm curious if more general result is known.
My attempts:
In the general case, my first thought was to use Hall's marriage theorem. We would need to prove that for any sets $A_1,\ldots, A_j \in \mathcal C^l$ there are at least $j$ sets $B\in \mathcal C^k$ that $B\in A_i$ for some $1\leq i\leq j$. This doesn't seem to be easy.
In some particular cases we can figure out the definition of $\varphi$.
- Consider the case $n=5$, $k=2$, $l=3$. We can treat numbers 1,2,3,4,5 modulo 5, so 6=1, 7=2 and so on. Now, if $A\in \mathcal C^3$ has three consecutive numbers (modulo 5) than we remove the middle one. For example $$\varphi(\{1,2,3\})=\{1,3\},\quad \varphi(\{4,5,1\})=\{4,1\}.$$ If not, there are two consecutive numbers in $A$ and an 'outlier' and then we remove this outlier. For example: $$\varphi(\{1,2,4\})=\{1,2\},\quad \varphi(\{5,1,3\})=\{5,1\}.$$ It's easy to see that its injection.
- Consider the case $n=6$, $k=3$, $l=4$ (as in the above mentioned olympiad problem). Then we can divide $\mathcal C^4$ with respect to the fact if the set contains $6$. Let $\varphi$ be from the previous case. We can define the function $\psi\colon \mathcal C^4\to \mathcal C^3$ as follows: $$ \psi(A) = \begin{cases} \varphi(A\setminus\{6\})\cup\{6\}; & 6\in A \\ A\setminus\{a'\};& A = [5]\setminus\{a\} \end{cases} $$ where $a'$ is the successor of $a$ in a 'cyclic set' $[5]$, that is $a' = 1+ (a\bmod 5)$.
I'm wondering if there is a better, more general way to define such functions, without having to play as above or if there is a theorem that it's always true.
The last idea is to use induction, in a similar manner as I did in the two particular cases above. Unfortunately, I need to know that the conclusion is true for $l=n-k$.
