$f$ isn't necessarily bijective but still $f^{-1}$ shows up

If $A$ is compact, is then $f(A)$ compact?

The answer here by David Mitra uses $$f^{-1}$$, however we only know $$f$$ is continuous in its domain, so how do we come up with the inverse?

Its not mentioned to be strictly montone either. I do think it's a stupid question but I really want to know.

• If $A$ is a set and $f:X\to Y$ a function, then $f^{-1}(A)$ is always defined and is by definition $\{x\in X\mid f(x)\in A\}$.
– Surb
Aug 1 '20 at 7:42

For any function $$f: X \to Y$$ and any set $$A \subseteq Y$$ we define $$f^{-1}(A)$$ as $$\{x \in X: f(x) \in A\}$$. This is called the inverse image of $$A$$ under $$f$$.

• Thanks, I got confused
– user732848
Aug 1 '20 at 7:44
• @Shamim It may in fact be confusing. The symbol $f^{-1}$ has two interpretations: (1) For each subset $A \subset Y$, $f^{-1}(A)$ denotes the inverse image (also called preimage) of $A$ which is a subset of $X$. Thus you may regard $f^{-1}$ as function from the power set of $Y$ to the power set of $X$. (2) If $f : X \to Y$ is bijective, then it is standard to write $f^{-1} : Y \to X$ for the inverse bijection. Usually it is absolutely clear from the context which interpretation is meant. Aug 1 '20 at 8:08
• Yes, I knew that definition beforehand but forgot right then, so I got confused.
– user732848
Aug 1 '20 at 9:04

In your given link, $$f^{-1}$$ is only used just for preimage of the set $$A$$. So, in this case, $$f^{-1}$$ need to be just a relation! You can say , bijective is necessary, if $$f^{-1}$$ is used as $$f^{-1} (a)$$ for $$a\in Im(f)$$, which is a inverse function of $$f$$.