Conditions on the Pre-Image

Question

Preimage: the pre-image of a set $S\in\mathbb{R^m}$ under a mapping $f:\mathbb{R^n}\rightarrow\mathbb{R^m}$ is the set $f^{-1}(S)=\{\vec{x}\in\mathbb{R^n}:f(\vec{x})\in S\}$

My question is, does the mapping of $f$ have to be continuous? It seems intuitive, but the books I have don't state this explicitly.

I'm not sure. I was just wondering if the definition should say $f$ is continuous. — , Jun 16 '18 at 03:12
Do you mean continuous in the sense of a topology? If so, note that invertibility does not guarantee continuity. Related: https://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous — SystematicDisintegration, Jun 16 '18 at 03:12
Continuity is not required to define the preimage; only invertibility is demanded. — SystematicDisintegration, Jun 16 '18 at 03:13
Note: Invertibility means that $f$ has an inverse $f^{-1}$, i.e. $f^{-1}$ is also a function. Invertibility has nothing to do with the notion of pre-image. — mcmayer, Jun 16 '18 at 03:22
The definition is just a definition, it places no other restrictions on $f$. To answer your question, no it does not need to be continuous. — copper.hat, Jun 16 '18 at 03:43

score 1 · Accepted Answer · answered Jun 16 '18 at 03:18

No, the function does not have to be continuous.

Look at the general setting where you have a function $f: A \rightarrow B$. $A$ and $B$ are sets. The pre-image of a set $B^\prime \subset B$ is $\{a\in A: f(a)\in B^\prime\}$, i.e. all elements from A that map into something in $B^\prime$.

If you know the function you can find the pre-image for any $B^\prime\subset B$ by figuring out what $\{a\in A: f(a)\in B^\prime\}$ is. You do not need to use any special property of $f$ to find a pre-image.

Conditions on the Pre-Image

1 Answers1