Linked Questions

7
votes
3answers
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 ...
7
votes
3answers
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?
8
votes
2answers
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.
0
votes
1answer
68 views

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 ...
1
vote
2answers
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 $X$ ...
1
vote
1answer
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?
9
votes
2answers
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'...
8
votes
2answers
3k views

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 ...
13
votes
1answer
2k views

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 ...
1
vote
2answers
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 ...
2
votes
3answers
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 ...
0
votes
1answer
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 $y_0 = f(x_0)$?
3
votes
2answers
93 views

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, $Y$ ...
1
vote
2answers
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 ...
1
vote
2answers
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?
1
vote
2answers
55 views

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 $f^{-1}$ ...
0
votes
0answers
117 views

does inverse of continuous function is also continuous? [duplicate]

Suppose that $f:R\rightarrow R $ is continuous and does have an inverse function. How can I prove that $f^{-1}$ is also continuous?
0
votes
1answer
69 views

Continuous function whose inverse is not continuous [duplicate]

Suppose $f:\mathbb{R}^n\to\mathbb{R}^n$ is bijective and continuous. Is it possible that $f^{-1}$ is not continuous? I can prove that for $n=1$ it is not possible, i.e. if $f:\mathbb{R} \to \mathbb{R}...
0
votes
0answers
73 views

Suppose $f$ is a continous bijection from compact metric space $X$ onto a metric space $Y$ . Then $f^{-1}$ is continuous mapping of $Y$ onto $X$

I read this theorem in rudin's book: Suppose $f$ is a continous bijection from compact metric space $X$ onto a metric space $Y$ . Then $f^{-1}$ is continuous mapping of $Y$ onto $X$ I have some ...
0
votes
0answers
67 views

'Reversing the implication' on the delta-epsilon definition of a limit

Is this logically valid? And if so, could someone please explain to me the name of the logical construct I'm performing (e.g. converse, contrapositive) as well as its mechanics? Please note that my ...
1
vote
0answers
62 views

Inverse of a continuous function is continuous

To prove for an open set, the inverse of a continuous function is continuous. The proof is, Let $F:\mathbb{R}\rightarrow\mathbb{R}$ be continuous. It is sufficient that $F^{-1}$ is the inverse of $F$...
1
vote
0answers
42 views

Function from Reals to Reals and has an inverset

We have seen in topology, that if a function is continuous and has an inverse, its inverse is not necessarily continuous. However the examples we had were from higher dimensions. So I've been thinking ...
1
vote
1answer
37 views

Conditions on the Pre-Image

Preimage: the pre-image of a set $S\in\mathbb{R^m}$ under a mapping $f:\mathbb{R^n}\rightarrow\mathbb{R^m}$ is the set $f^{-1}(S)=\{\vec{x}\in\mathbb{R^n}:f(\vec{x})\in S\}$ My question is, does the ...