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Given two finite sets $A$ and $B$ with $|A|<|B|$ There are more functions from $B$ to $A$ than from $A$ to $B$ except when $|A|=1$ or $|A|=2,|B|=3,4$. See here for proof.

It is also true there are more surjections from $B$ to $A$ than injections from $A$ to $B$.

One way to prove this is as follows: Create a bipartite graph in which he vertices in one part are the injections from $A$ to $B$ and the vertices in the other part are the surjections from $B$ to $A$. Connect two vertices if and only if they are inverses (on the appropriate sides) of each other.

injective functions have $(|A|)^{|B|-|A|}$ left inverses. On the other hand the number of right inverses a surjective function can have depends on the function itself. It is the product of the order of the pullbacks of all the singletons in the image. So then it is the product of $A$ numbers which add $B$ and hence by AM-GM it is bounded by $(\frac{|B|}{|A|})^{|A|}$ Therefore we have $I(|A|)^{|B|-|A|}\leq S(\frac{|B|}{|A|})^{|A|}$ where $I$ is the number of injective functions and $S$ of surjective functions. This is because the sum of the degrees of each of the two parts of a bipartite graph is equal.

Therefore $\frac{S}{I}\leq\frac{|A|^{|B-|A|} |A|^{|A|}}{|B|^{|A|}}=\frac{|A|^{|B|}}{|B|^{|A|}}$.

On the other the functions from $B$ to $A$ and from $A$ to $B$ are in ratio $\frac{|A|^{|B|}}{|B|^{|A|}}$.

Therefore the number of injective functions are a smaller percentage of the functions from $A$ to $B$ than the surjective functions are from $B$ to $A$.

This leads me to believe being surjective is "easier" than being injective.

Follow up questions:Let $|A|<|B|$ for arbitrary sets $A$ and $B$.

Is there a way to prove there are more injective functions from $A$ to $B$ than surjective functions from $B$ to $A$?

Is there a way to prove $|A^B|\geq |B^A|$?

Is there a way to express the fact that surjections are "easier" to find in infinite sets? Would this be true? Clearly we can't use the same argument.

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  • $\begingroup$ Great, thanks for the edit. $\endgroup$ – Hayden Jan 11 '15 at 20:59
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There's an easy way to show the inequality for infinite sets. Cardinal arithmetics!

Recall that for an infinite cardinal $\kappa$ it is true that $\kappa^\kappa=2^\kappa$ (since $2^\kappa\leq\kappa^\kappa\leq(2^\kappa)^\kappa=2^{\kappa\cdot\kappa}=2^\kappa$).

Assuming that $|A|>2$ we immediately get that $|A^B|=2^{|B|}$, and $|B^A|\leq|B^B|=2^{|B|}$, so we indeed have the inequality wanted. If you take, for example $A=\Bbb N$ and $B=\Bbb R$, then you also get a sharp inequality there, since $A^B$ has size $2^{2^{\aleph_0}}$ whereas $B^A$ has size $2^{\aleph_0}$.

(Note that for "arbitrary sets" it's hard to prove there are more injective functions from $A$ to $B$ since if $A$ is empty there are no surjective functions, and there is one injection; and if $A$ is a singleton and $B$ is not, then there is only one surjection, but many injections; and if $A$ has at least two elements then there are $2^{|B|}$ surjections, but not necessarily as many injections. So we can't say much about "arbitrary sets".)

All in all, I'd say that it's "easier" to find surjections exactly by expressing this sort of cardinal inequality. The set of surjections from $B$ to $A$ is not smaller in cardinality than the set of injections from $A$ into $B$ (assuming, of course, $A$ has at least two elements, and $|A|\leq|B|$).

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