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Suppose I have a function $f: X \rightarrow X$, a finite set $S\subset X$ and $|S'| \neq |S|$ for $f(S) = S'$. Is there a special name for such a function $f$?

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Not $1-1. \ \ \ \ $ – P.. Mar 2 at 13:59
I thought this was going to be about infinite sets! Does anyone know the name for it in that case? – Tara B Mar 2 at 14:02
Ah, no, $X$ is finite as well – Confusion Mar 2 at 14:03
Yes, that doesn't really matter. What I thought you were going to be asking for (from the title) is the name for a function $f: X\to X$ where $|f(X)|<|X|$. If $X$ is infinite, that is a much stronger property than 'not injective'. – Tara B Mar 2 at 14:06
I realised that the comments on another question are a bad place to ask a question, so I've now asked it here. – Tara B Mar 2 at 14:42
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3 Answers

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If you want to specifically mention that the image of $S$ under $f$ is smaller than $S$, you could say that $f$ is "not injective on $S$". Of course $f$ is also not injective at all, as the other answers say, but that is weaker.

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+1, good answer, in fact the formulation of the question appears to require a name that refers to both $f$ and $S$. – Andreas Caranti Mar 2 at 14:10
@AndreasCaranti: Thanks. I do think this is probably what the OP was looking for. – Tara B Mar 2 at 14:11
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Then you must have $|S'| < |S|$, and you are talking of a function which is not injective.

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Yes: in that case it must be that $|S'|<|S|$, and therefore $f$ is not injective.

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Not literally the same answers, but we came close ;-) – Andreas Caranti Mar 2 at 14:00
@Andreas: Hard to get much closer! – Brian M. Scott Mar 2 at 14:00
So given my edit, 'not injective and not surjective', but that's all – Confusion Mar 2 at 14:01
@Confusion: If $X$ is infinite, your function could be surjective and still have the stated property. E.g., $$f:\Bbb N\to\Bbb N:n\mapsto\left\lfloor\frac{n}2\right\rfloor\;.$$ – Brian M. Scott Mar 2 at 14:02

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