# Preimage of a singleton set.

My question is aroused by this article;

"By definition of a function, the image of an element x of the domain is always a single element y of the codomain. Conversely, though, the preimage of a singleton set (a set with exactly one element) may in general contain any number of elements. For example, if f(x) = 7 (the constant function taking value 7), then the preimage of {5} is the empty set but the preimage of {7} is the entire domain."

Here,I can understand that the preimage of the singleton set is the entire domain.But if so,then how does the inverse of a singleton function be a "function" according to the function definition as there exists more than one element map singleton set to preimage.

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## 3 Answers

For a function to have an inverse, it has to be injective--that is, distinct domain elements must be taken to distinct codomain elements. In such a case, preimages of singletons will either be singletons or the empty set. However, preimages are defined even when a function doesn't have an inverse.

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A function need not have an inverse to consider the preimage.

As you said the preimage of a singleton set under a constant function can be the whole domain, so there is no well defined inverse function here.

The preimage, however, is simply a set $f^{-1}(E) = \{x : f(x) \in E \}$, which always makes sense.

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You use the phrase, "singleton function" --- I don't know what you mean by that. In any event, in general the inverse of a function isn't a function, so there is no problem. Alternatively, if you refuse to use the word "inverse" to describe something that isn't a function, then there are functions that don't have an inverse.

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