# Does the following inverse function really exist?

Let $f : A \rightarrow B$ be the function $f(a) = 0 \forall a \in A$ with A = $\Re$and B = $[0,1]$

In the solution of my exercise, it is written $f^{-1} (B) = \Re$

I thought that for an inverse to exist for a given function, the function needs to be surjective (and injective too). If this is the case, all elements of the codomain (in this case B) should be mapped to from an element in A. But in this case, the function only maps to 0, not the other elements in B, how is this possible ?

• $f^{-1}(B)$ denotes the preimage and not the inverse function. – TheGeekGreek Sep 13 '17 at 22:17
• I'd like to clarify a few things—perhaps they are relevant to you, perhaps they are relevant to the person who wrote the exercise. There's a redundant amount of precision here: $f:A\to B$ means “$f$ is a function under which every element (yes, all of them) of $A$ is mapped to some element of $B$.” If you see $f:x\mapsto y$ (such as $x\mapsto x^2$), it means “$f$ is a function that maps $x$ to $y$” (i.e., $y=f(x)$). That whole first line could be compressed to $$f:\Re\to\{0\}$$ but that wouldn't make as good an exercise I suppose. – gen-ℤ ready to perish Sep 14 '17 at 2:31

$f^{-1}(B)$, where $B$ is a set, is different than $f^{-1}(b), b \in B$
The definition of $f^{-1}(B)$ is $\{x \in A| f(x) \in B\}$. So it is the set of values in the domain that get mapped into the set $B$. In your case, each value in A is mapped to 0, which is inside $B=[0,1]$