Domain and preimage of a relation/function

Question

The preimage of a function $f:X \rightarrow Y$ is defined as $Img^{-1} f = \{ x \in X : f(x) \in Y \} $, which is equivalent to $\{ x \in X : \exists y \in Y : f(x) = y \}$. This last definition is the same of the domain of a function. Are preimage and domain always equivalent? Or is there a case where we have an element in the domain and not in the image?

Answer 1

The preimage is usually defined for subsets of the codomain. That is, given a function $f : X \to Y$ and a subset $V \subseteq Y$, the preimage of $V$ under $f$ is defined by $$f^{-1}[V] = \{ x \in X : f(x) \in V \}$$ The set in your question is the preimage of the whole codomain—that is, using your notation, we have $\mathrm{Img}^{-1}(f) = f^{-1}[Y]$. Since part of the definition of a function is that its values should be elements of the codomain, we inevitably get $f^{-1}[Y]=X$.

So yes, the preimage of the whole codomain of $f$ is the domain of $f$, but the notion of 'preimage' is defined more generally for arbitrary subsets of the codomain.