0

Say $f:X\to Y$, $A\subseteq X$ and $B\subseteq Y$. This answer proves $f(f^{-1}(B))=f(X)\cap B$ (n.b.: the author of that answer uses different notation from me). Is there an analogous identity for $f^{-1}(f(A))$? I initially conjectured $f^{-1}(f(A))=f^{-1}(Y)\cup A$, but that conjecture fails since $f^{-1}(Y)$ is just $X$. I then considered the conjecture $f^{-1}(f(A))\cap f^{-1}(Y)=A$, but this conjecture fails for the same reason. I think there is no such identity, but I can't justify why.

Edit: I am already familiar with the relations $f(f^{-1}(B))\subseteq B$ (equality iff $f$ surjective) and $f^{-1}(f(A))\supseteq A$ (equality iff $f$ injective).

xFioraMstr18
  • 867

1 Answers1

1

Let $X$ and $Y$ be sets. Fix $f:X\to Y$.

Remark. For each $A\subseteq X$, $$A\subseteq f^{-1}(f(A))=A\cup\{x\in\mathrm{domain}(f):(\exists a\in A)[f(x)=f(a)]\}.$$ Furthermore, equality holds if and only if $f$ is injective.

Remark. For each $B\subseteq Y$, $$ B\cap\mathrm{range}(f)=f(f^{-1}(B))\subseteq B.$$ Furthermore, equality holds if and only if $f$ is surjective.

Alberto Takase
  • 3,511
  • Okay, this identity $f^{-1}(f(A))=A\cup{x\in\mathrm{domain}(f):(\exists a\in A)[f(x)=f(a)]}$ is true, but it seems cheating; we already have $f^{-1}(f(A))={x\in\mathrm{domain}(f):(\exists a\in A)[f(x)=f(a)]}$ and $A\subseteq{x\in\mathrm{domain}(f):(\exists a\in A)[f(x)=f(a)]}$, so the union is a trivial union. Is there any nontrivial identity? – xFioraMstr18 May 10 '18 at 11:55
  • 1
    You seem more interested in brevity over clarity. Then $f^{-1}(f(A))$ is (from my experience) the most compact (standard) notation. Feel free to invent a more compact notation. I thought you were interested in an identity which does not use the preimage $f^{-1}$. Can I ask, what exactly are you seeking in an identity? (trivial/nontrivial is subjective so I cannot help you there) – Alberto Takase May 11 '18 at 15:28