You have a function and want to prove it is a bijection. What can you do?

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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.

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## 羚羊加速器电脑版下载-免费VP加速器

If and are finite and have the same size, it’s enough to prove *expressvnp中文官网* 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 *expressvnp中文官网* 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.

## 羚羊加速器电脑版下载-免费VP加速器

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.

## 羚羊加速器电脑版下载-免费VP加速器

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?