# Is a Function From $X$ "into"$Y$ Injective?

I am working my way through Naive Set Theory by Halmos, and I am trying to solve the exercise:

Suppose that $f$ is a mapping from $X$ into $Y$ and $g$ is a mapping from $Y$ into $X$. Prove that there exist subsets $A$ and $B$ of $X$ and $Y$ respectively, such that $f(A) = B$ and $g(Y - B) = X-A$.

This result can be used to give a proof of the Schroder-Bernstein theorem that looks quite different from the one above.

Given the context, it seems that saying $f$ maps $X$ "into" $Y$ implies injectivity. However, I have read conflicting opinions on whether "into" implies injectivity or whether it just means that $Y$ is the co-domain of $f$. Was "into" common slang for an injection in older texts?

• For some basic information about writing math on MSE see here, here, here and here.
– user409521
Apr 8, 2017 at 0:44
• An "into" or "one-to-one" function is also called an "injection"
– user409521
Apr 8, 2017 at 0:45
• Yes, "into" is one of those words that we have to be extra careful about. Most often it is used to mean one-to-one or injective -- but it's better to double-check with definitions earlier in the same book. Apr 8, 2017 at 1:02
• I'm not sure whether it is "widely accepted" or not and I agree with @zipirovich that it is rather ambiguous and can lead to confusion if not clear from context. For example, bijective functions are better described as "one-to-one" and "onto" as opposed to "into" and "onto".
– user409521
Apr 8, 2017 at 1:05
• Personally I have never seen into used to mean injective, so at least it is not standard. Apr 8, 2017 at 6:19

A mapping "from $X$ into $Y$" does not imply injectivity. The quoted theorem, sometimes called the Banach mapping theorem or Banach decomposition theorem, is valid for arbitrary finctions $f:X\to Y$ and $g:Y\to X,$ not necessarily injective. Of course, only the special case of injective functions is needed for the proof of the Schröder–Bernstein theorem. On the other hand, the "Banach mapping theorem" for arbitrary functions is easy enough to prove, and restricting it to injective functions does not lead to any simplification that I know of.
The conclusion $$\exists A \exists B\ [f(A)=B\text{ and }g(Y-B)=X-A]$$ can be rewritten as $$\exists A\ X-g(Y-f(A))=A,$$ that is, $A$ is a fixed point for a certain mapping from $\mathcal P(X)$ to $\mathcal P(X).$