Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X$ is some connected topological space, $I$ is an interval and $Y$ is some topological space. Let $g: X\to I$ be a continuous and surjective function. Let $f$ be a function $I \to Y$. If $f \circ g$ is continuous, must $f$ be continuous?

To prove this is suffices to prove that if $x \in I$ and $x_n$ converges to $x$ from either below or above then $f(x_n)$ converges to $f(x)$. Let us assume wlog that $x_n \geq x$ for all $n$. It would be enough to find $y$ and $y_n$ such that $g(y_n)=x_n$, $g(y)=x$ and $y_n \to y$.

There is some $y \in X$ such that $f(y)=x$. In addition we can pick $y$ so that every open set $ U \ni y$ contains a $z$ with $g(z)>x$. (Follows from connectedness.) If $X$ were first-countable I think this would imply the existence of the desired sequence $y_n$. But I cannot see how to conclude this in general.

share|improve this question
2  
Is $I$ a closed or an open interval? –  Nate Eldredge Nov 21 '12 at 15:12
    
Any interval; open, closed or half-closed. Of course if it turns out that the answer depends on the type of interval I would find that interesting. –  Johan Nov 21 '12 at 15:56

1 Answer 1

up vote 4 down vote accepted
+100

$f$ need not be continuous.

Let $X = Y$ be the topologist's sine curve, i.e. $$X = Y = \{(x, \sin(1/x)) : 0 < x \le 1\} \cup (0,0) \subset \mathbb{R}^2$$ with the Euclidean topology. It is well known and easy to check that $X$ is connected. Take $I = [0,1]$ and define $g : X \to I$ by $g(x,y) = x$; then $g$ is a continuous bijection. Let $f = g^{-1}$, i.e. $$f(x) = \begin{cases} (x, \sin(1/x)), & 0 < x \le 1 \\ (0,0), & x=0. \end{cases}$$ Then $f \circ g$ is the identity map on $X$ which is continuous, but $f$ is not continuous.

An example with $I$ an open interval can be constructed similarly (for instance, take $I=(-1,1)$ and just reflect this picture about the $y$ axis).

share|improve this answer
    
Thank you, I realize now what went wrong in my supposed proof for first-countable spaces. –  Johan Nov 21 '12 at 16:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.