A classical version of the inverse function theorem for 1D asserts the following.

Suppose that is an invertible function with inverse . If is differentiable at with and is continuous at , then is differentiable at and

Very often, some of the above hypotheses for is replaced by, say, is continuous “in a neighborhood” and one-to-one. But in the proof, while the one-to-one property immediately gives the existence of the inverse function , the “non-local” continuity yields the continuity of the inverse . Here we are interested in the “local” continuity/differentiablity.

In this note, we show by constructing counter-examples that in general the continuity of at cannot be dropped. In the sequel, we denote by the set set of positive integers, namely .

**1. The case of continuity**

We start with the simplest case, namely if we only assume that is continuous at . In this scenario, the following example due to R.E. Megginson is well-known. Consider the function defined by

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