Suppose we have sets $A$ and $B$ and a function $f : A \to B$ (that is, $f$’s domain is $A$ and its codomain is $B$). Suppose there is another function $g : B \to A$ such that $g(f(a)) = a$ for every $a \in A$. Is $f$ necessarily a bijection? That is, does $f$ necessarily match up each element of $A$ with a unique element of $B$ and vice versa? Or put yet another way, is $f$ necessarily invertible?

• If yes, prove it!
• If no, provide a counterexample! For bonus points, what additional assumptions could we impose to make it true? Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 12 Responses to Test your intuition: bijections

1. Sylvain B. says:

There is no restriction regarding the cardinality of domain and codomain?

• Brent says:

No, as stated they can be any sets. But if you think it would help/make a difference to restrict their cardinality then that would be an interesting observation.

2. designerspaces says:

f : Nat -> Int is the usual inclusion and g : Int -> Nat is the absolute value. Then we have:
g (f n_Nat) = |n _ Int| = n_Nat

General comments which Brent well knows: In this case g is a left inverse and such a map f is called a split monomorphism (it makes sense in an arbitrary category to talk functions with left inverses). We have the implication that split mono => mono but not conversely. In Set we have:

injective monomorphism split monomorphism
In general in a category it is the case that being a mono + epi does not imply being an isomorphism but it is the case that split mono + epi iso (this is straight forward because the pre-composition function for such an f : X -> Y, Hom (Y, Z) -> Hom (X, Z) is a bijection).

One additional heavy-handed way to make it true: Have A = B and both finite. This then follows by the pigeonhole principle. By the above we can also ask that f additionally be surjective!

3. Michael Paul Goldenberg says:

Perhaps I’m being naive, but wouldn’t it suffice to have the co-domain contain elements that aren’t hit by f? Then g would return every element in the co-domain to a in A, but there would be elements in B that aren’t equal to f(a) for any a; that would mean that g(b) might not be defined for one or more elements in B. No bijection in that case. To guarantee a bijection, f would have to be injective and surjective for B,

• Brent says:

Not naive at all, this is exactly the right kind of idea. However, note that by saying $g : B \to A$ I am implicitly claiming that g is in fact defined on all the elements of B. But f still wouldn’t be a bijection in that case.

4. Buddha Buck says:

Let $A$ be a singleton set, and $B$ be any set with 2 or more elements. Then any pair of functions $f:A\to B, g:B\to A$ form a counter-example. There is only one possible $g$, which maps all elements of $B$ to the single element in $A$, so no matter the value of $f(a)$, you will have $g(f(a)) = a$. Yet there is clearly no bijection between $A, B$.

• Brent says:

That’s a nice, simple counterexample! Though (depending on your comfort level with functions out of an empty set) you could make it even simpler by making A empty and B a singleton set. =)

• Matt says:

I’m fine with functions out of an empty set. The existence of g to an empty set on the other hand…

• Brent says:

Oh! Of course, that doesn’t work, how silly of me.

• blaisepascal2014 says:

How about $A = \{\emptyset\}, B = \{\emptyset,A\}$? I am not sure it is possible to get simpler than that.

• Brent says:

I think your original formulation (which didn’t need to mention specific elements of A or B at all) was simpler!