9

If I there exists an injection $\phi: S_1 \to S_2$ and a surjection $\tau: S_1 \to S_2$, does there necessarily exist a bijection between sets $S_1$ and $S_2$?

I'd like this to be true, but I don't see a way to construct a bijection directly from $\phi$ and $\tau$.

hchau
  • 343
  • Depends of AC.. – Gaston Burrull Feb 26 '13 at 01:19
  • 1
    @hchau, could you tell us the level that you're learning at? Depending on if you're learning elementary discrete maths, or axiomatic set theory, it can make a big difference to what form your answer takes. – Dan Rust Feb 26 '13 at 01:23
  • Oh, so I haven't actually taken an undergraduate set theory course yet, but this question just popped into my mind, so I thought I'd ask it here. – hchau Feb 26 '13 at 01:32
  • I have actually answered this question at least once on this site (and I believe that it was twice or thrice). – Asaf Karagila Feb 26 '13 at 05:54
  • Yes, I have found four copies: http://math.stackexchange.com/questions/46168/proof-of-a-cantor-bernstein-like-theorem and http://math.stackexchange.com/questions/108810/does-a-injective-function-f-a-to-b-and-surjective-function-g-a-to-b-impl/ and http://math.stackexchange.com/questions/176972/is-there-a-cantor-schroder-bernstein-statement-about-surjective-maps/ and http://math.stackexchange.com/questions/220617/sur-in-bijections-and-cardinality/ – Asaf Karagila Feb 26 '13 at 06:06

2 Answers2

9

The statement that if there is a surjection from $A$ to $B$, then there is an injection from $B$ to $A$ is known as the Partition principle. It is a consequence of the axiom of choice, and it’s not known whether it is equivalent to the axiom of choice. Given the partition principle, the existence of both an injection and a surjection from $A$ to $B$ implies the existence of injections from $A$ to $B$ and $B$ to $A$, and the Schröder-Bernstein theorem then implies that there is a bijection from $A$ to $B$.

Brian M. Scott
  • 616,228
4

Cantor-Bernstein-Schroeder Theorem

And the fact that, if there exists a surjection from A to B, then there exists an injection from B to A.

Jim
  • 30,682
Aloizio Macedo
  • 34,292
  • 4
    Assuming the axiom of choice. Which really should be mentioned in this case, since the Schröder-Bernstein theorem itself does not require AC. – Brian M. Scott Feb 26 '13 at 01:07
  • 2
    Strictly speaking, a surjection from A to B only implies an injection from B to A if you have some form of the axiom of choice at your disposal. – Dan Rust Feb 26 '13 at 01:08