You have a function and want to prove it is a bijection. What can you do?
By the book
A bijection is defined as a function which is both one-to-one and onto. So prove that is one-to-one, and prove that it is onto.
This is straightforward, and it’s what I would expect the students in my Discrete Math class to do, but in my experience it’s actually not used all that much. One of the following methods usually ends up being easier in practice.
If and are finite and have the same size, it’s enough to prove either that is one-to-one, or that is onto. A one-to-one function between two finite sets of the same size must also be onto, and vice versa. (Of course, if and don’t have the same size, then there can’t possibly be a bijection between them in the first place.)
Intuitively, this makes sense: on the one hand, in order for to be onto, it “can’t afford” to send multiple elements of to the same element of , because then it won’t have enough to cover every element of . So it must be one-to-one. Likewise, in order to be one-to-one, it can’t afford to miss any elements of , because then the elements of have to “squeeze” into fewer elements of , and some of them are bound to end up mapping to the same element of . So it must be onto.
However, this is actually kind of tricky to formally prove! Note that the definition of “ and have the same size” is that there exists some bijection . A proof has to start with a one-to-one (or onto) function , and some completely unrelated bijection , and somehow prove that is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when and are infinite: a one-to-one function between two infinite sets of the same size need not be onto, or vice versa; we saw several examples in my previous post, such as defined by . Although tricky to come up with, the proof is cute and not too hard to understand once you see it; I think I may write about it in another post!
Note that we can even relax the condition on sizes a bit further: for example, it’s enough to prove that is one-to-one, and the finite size of is greater than or equal to the finite size of . The point is that being a one-to-one function implies that the size of is less than or equal to the size of , so in fact they have equal sizes.
One can also prove that is a bijection by showing that it has an inverse: a function such that and for all and . As we saw in my last post, these facts imply that is one-to-one and onto, and hence a bijection. And it really is necessary to prove both and : if only one of these hold then is called a left or right inverse, respectively (more generally, a one-sided inverse), but needs to have a full-fledged two-sided inverse in order to be a bijection.
…unless and are of the same finite size! In that case, it’s enough to show the existence of a one-sided inverse—say, a function such that . Then is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence is actually a two-sided inverse).
We must be careful, however: sometimes the reason for constructing a bijection in the first place is in order to show that and have the same size! This kind of thing is common in combinatorics. In that case one really must show a two-sided inverse, even when and are finite; otherwise you end up assuming what you are trying to prove.
By mutual injection?
I’ll leave you with one more to ponder. Suppose is one-to-one, and there is another function which is also one-to-one. We don’t assume anything in particular about the relationship between and . Are and necessarily bijections?
Cheeky example: S: Nat -> Nat is injective so take f = g = S!
By S do you mean the successor function? That’s a good example, not cheeky at all!
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I did mean successor, yes!
If is one-to-one, then . Similarly, if , then . If both are true, then and a bijection exists between , but there’s no guarantee that either are bijections.
If are finite, then (as you argued in “By Size”), both are bijections. But if they are infinite, then there are that are one-to-one, but not onto. Proof: Since is infinite, there exists a that is one-to-one but not onto, so if is one-to-one, then is one-to-one, but not onto.
This is one of my favorite ways to show . It usually isn’t hard to come up with the necessary injections, even if it is hard to find a bijection. It is easy to show that and are both one-to-one, It is harder to come up with an easy-to-describe bijection.
Indeed! I think I will probably write about this in an upcoming post. One fun thing is that the proof of the Schröder-Bernstein theorem is actually constructive, so in theory you can take any two injections and actually use them to construct a bijection. But indeed, there is no guarantee that the resulting bijection is easy to describe. I am still trying to work out how to describe the one generated by your pair of injections between and . It’s something like this: any number can be written in the form 2^2^…^m where is not a power of two, that is, a tower of zero or more 2’s with a final power of on top. (If is not a power of two then .) If or has prime factors other than 2 or 3, then send to . Otherwise, if is a power of two, send it to ; finally, if , send it to . It’s very non-obvious that this is a bijection (and I might have even gotten the description wrong)!
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