14k views

### Does continuous imply continuous inverse? [duplicate]

Possible Duplicate: Functions which are Continuous, but not Bicontinuous If $f$ is a continuous map from a subset of $\mathbb{R}^n$ to another subset of $\mathbb{R}^n$, must it have a continuous ...
10k views

### Is the inverse of a continuous bijective function also continuous? [duplicate]

Is the inverse of a continuous bijective function also continuous? How to prove it?
1k views

### Is there a bijective continuous function $f: \Bbb R \to \Bbb R$ whose inverse $f^{-1}$ is not continuous? [duplicate]

Looking at the definition of an homeomorphism, this question came to my mind.
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### Preimage of continuous one-to-one function on connected domain is not continuous. [duplicate]

I know that given $B$, a compact subset of $\mathbb{R}^n$, and $f : B \to \mathbb{R}^m$, a continuous injective (one-to-one) function, $f^{-1}$ is continuous on $f(B)$. (This true). I also know that ...
37 views

### A counter-example for continues functions on Metric spaces [duplicate]

Give a function $f : X \to Y$ such that $(X,d)$ and $(Y,p)$ are metric spaces, and $f$ is continues and one-to-one function and onto $Y$ but $f^{-1}$ is not continues ? I know that compactness of $X$ ...
34 views

### Will the inverse mapping also be continuous? [duplicate]

Suppose a map $f:A\to B$ is continuous and invertible. Will the inverse map $f^{-1}:B\to A$ be always continuous also?
3k views

### Example of a continuous function with a discontinuous inverse

What is an example of a function $f: \Bbb R^n \rightarrow \Bbb R^m$ such that $f$ is continuous and injective but that $f^{-1}$ is not continuous. Our professor teased us with the notion but I haven'...
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### Differentiable bijection $f:\mathbb{R} \to \mathbb{R}$ with nonzero derivative whose inverse is not differentiable

I had an exam today, and I was asked about the inverse function theorem, and the exact conditions and statement (as stated in Mathematical Analysis by VA Zorich): Let $X, Y \subset \mathbb{R}$ be open ...
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### Geometric Proofs of Calculus Theorems

I just started learning "rigorous" calculus, and I noticed that a lot of calculus theorems are rather obvious from the geometrical point of few. Some examples: 1. Prove that the derivative ...
1k views

### Invertible functions in $R^m$

The definition of an invertible function that my book (Apostol's Mathematical Analysis) gives is: A function $f:S \to\mathbf{R}^n$, where $S$ is open in $\mathbf{R}^n$, has a unique inverse if $f$ is ...
168 views

### Under what circumstances does continuity of a function imply continuity of its inverse?

I was working on a few proofs about topological equivalence, and I realized that things would be a lot easier if I could form some shortcut that exempted me from proving the continuity of both a ...
413 views

### Inverse of function continuous at a point must be continuous at that point?

If a function $y = f(x)$ is continuous at $x_0$. Suppose the function is invertible in a neigborhood of $x_0$. Is it true that the inverse function must be continuous at $y_0 = f(x_0)$?
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### Counterexamples to: If $f:X \to Y$ is continuous, $Y$ is compact, then $f^{-1}$ is continuous.

It is a theorem that if $f:X \to Y$ is a continuous bijection, $X$ is compact, then $g = f^{-1}$ is continuous. My professor asked us to find a counterexample to If $f:X \to Y$ is continuous, $Y$ ...
141 views

### The inverse of a continuous one-to-one function that is defined on a connected set is not always continuous

I am trying to find a function $f:B \subset \Bbb R^n \rightarrow \Bbb R^m$ for $B$ a connected set that is continuous, one-to-one where $f^{-1} = f(B) \rightarrow B$ is discontinuous. The hint I have ...
93 views

### Is the map, $f:(0,1)⊂ \mathbb{R}$ → $(1,∞)⊂ \mathbb{R}$ : $x ↦ 1/x$continuous?

I feel it is, but cannot prove why. Also is it bijective, and is its inverse continuous?
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### Can there exist a continuous 1-1 function $f$ such that $f^{-1}$ exists but is not continuous?

I'm doing some basic topology and analysis in the book "Tensor Analysis on Manifolds" by Bishop and Goldberg, and a homeomorphism is defined as a bijection, $f: X \to Y$, such that $f$ and $f^{-1}$ ...
Suppose that $f:R\rightarrow R$ is continuous and does have an inverse function. How can I prove that $f^{-1}$ is also continuous?