# Infinite cardinals $\alpha < \beta$ and “locally injective” function $f:\beta \to \alpha$ [closed]

Are there infinite cardinals $$\alpha < \beta$$ and a function $$f:\beta\to\alpha$$ with the following property?

For any $$S\subseteq \beta$$ with $$|S| =\alpha$$ the restriction $$f|_S:S\to \alpha$$ is injective.

## closed as off-topic by Emil Jeřábek, Joseph Van Name, Pace Nielsen, Jan-Christoph Schlage-Puchta, Ben BarberFeb 4 at 12:34

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Emil Jeřábek, Pace Nielsen, Ben Barber
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Joseph Van Name, Jan-Christoph Schlage-Puchta
If this question can be reworded to fit the rules in the help center, please edit the question.

• meta.mathoverflow.net/questions/2781/decluttering-mathoverflow – Mere Scribe Feb 1 at 10:43
• @Mere: The OP cannot delete a question if it has (1) an answer with an upvote (let alone 4 and a check mark), or (2) two answers. It will not be removed by the Roomba script either. So it really just waits three people to vote to delete it (which may as well include the OP, of course) or for the OP to unaccept the answer, so Joel can delete it, so they can delete the question. – Asaf Karagila Feb 5 at 10:57
• First, apologies for the bad question. Then: wouldn't it be disrespectful to unaccept @JDHamkins' answer and ask him to delete it so I can delete this question? I would like to remove it, but I don't want to offend anyone. – Dominic van der Zypen Feb 5 at 18:10

This is impossible. Since $$f$$ cannot be injective, as $$\alpha<\beta$$, there will be two points with the same $$f$$ value, $$f(x)=f(y)$$, and if we place those two points $$x$$, $$y$$ into an $$S$$ of size $$\alpha$$, we cannot have $$f\upharpoonright S$$ injective.
• Basically, the point is that injectivity is itself a highly local notion, since $f$ is injective if and only if the restriction of $f$ to all two-element subsets of the domain is injective. – Joel David Hamkins Jan 31 at 13:26