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.

$f:X\rightarrow Y$ is a continuous surjection Then

  1. If $V$ is open, does this imply $f(V)$ is open?

  2. If $F$ is closed, does this imply $f(F)$ is closed?

  3. If $A$ is an infinite subset, does this imply that $f(A)$ is so in $Y$?

Thank you for help.

share|improve this question
add comment

6 Answers

up vote 6 down vote accepted

None of these is true.

Let $\Bbb S$ be the Sierpiński space, i.e. $\Bbb S=\{0,1\}$ where the open sets are $\emptyset,\{1\}$ and $\{0,1\}$.

Define $f:\Bbb R\to\Bbb S$ as follows: $$f(x)=\begin{cases}0;&x\leq 0\\1;&x>0\end{cases}$$

This is a continuous surjection, since $f^{-1}(\{1\})$ is open, but none of the properties holds:

  • $f((-\infty,0))=\{0\}$ is not open
  • $f([1,\infty))=\{1\}$ is not closed
  • the image of $\Bbb R$ is finite.
share|improve this answer
    
could you check once more about $f$? –  Bunuelian Trick Jul 12 '12 at 11:22
    
Point 2 seems false, since $f(0) = 0$. $f([1,\infty))=\{1\}$, though. –  Arthur Jul 12 '12 at 11:26
    
@Arthur: Thanks, modified it. It should be ok now. –  Dejan Govc Jul 12 '12 at 11:27
    
@Dejan, you need to redefine $f$ I think typo. –  Bunuelian Trick Jul 12 '12 at 11:27
1  
@TaxiDriver: thanks. Well, $\mathbb S$ is one of the spaces I really like to think about (and seemed appropriate) so I just had to try it. =) –  Dejan Govc May 31 '13 at 21:47
show 2 more comments

$3$ is false as you can construct functions that are locally constant and choose $X$ disconnected (each connected component infinite) for a counterexample.

share|improve this answer
    
Let $X=(0,1)\cup (2,3)$ in usual subspace topology, then locally constant functions are surjective? what about $1,2$ –  Bunuelian Trick Jul 12 '12 at 11:14
    
Let $Y$ be the two point set with any topology! They will be because the inverse image of each point is either $(0,1)$ or $(2,3)$. So continuity holds. So does surjectivity! –  Aneesh Karthik C Jul 12 '12 at 11:15
    
okay got it :)) –  Bunuelian Trick Jul 12 '12 at 11:17
add comment

1 is false. Consider $f(x) = |x| \sin x$, then $f((-\pi, \pi)) = [-\pi/2, \pi/2]$.

share|improve this answer
    
thank you :)))))))))) –  Bunuelian Trick Jul 12 '12 at 11:17
add comment

All three are false.

  1. $X=I$, $Y=S^1$ and $f$ glues the ends of $I$ together. Then any open subset $(a,1]\subseteq I$ is mapped to a non-open set.

  2. Let $X=Y$ as sets, where $X$ has the discret and $Y$ has the trivial topology. Any subset of $X$ is closed but not necessarily in $Y$.

  3. $X=Y\coprod Y$ with $Y$ infinite and $f$ the identity on one component and constant on the other one.

share|improve this answer
add comment

1 and 2 are false. Here are counterexamples for them:

  1. Let $X= [0,1)$ with the subspace topology and let $Y = S^1$ with the subspace topology. Let $f : X \rightarrow Y$ be the map defined by $$f(t) = (\cos 2 \pi t, \sin 2\pi t).$$ Then it is easy to see that not only is $f$ continuous, but that it is also bijective and hence surjective. I claim that $f$ is not an open map. Look at $V = [0,\frac{1}{4})$ that is open in $X$. Then we see that there is no $U$ open in $\Bbb{R}^2$ such that $f(0) \in U \cap S^{1} \subseteq f(V)$. It follows that $f$ is not open.

  2. Consider $\Bbb{R}$ with the Euclidean topology. Let $\Bbb{R}/\sim$ be the quotient space obtained by saying that $x \sim y$ iff $ x -y \in \Bbb{Q}$. Then it is not hard to see that the quotient topology on $\Bbb{R}/\sim$ is trivial. Furthermore $\Bbb{R}/\sim$ is uncountable. Now let $\pi $ be the canonical projection of $\Bbb{R}$ onto the quotient. Then $\Bbb{Z}$ is closed in $\Bbb{R}$ but $\pi(\Bbb{Z}) = [0]$ that is not closed in $\Bbb{R}/\sim$.

share|improve this answer
add comment

Here are two definitions which may be helpful for you:

If $1$ is true, we call $f$ is an open function; If $2$ is true, we call $f$ is a closed function. All these functions are different from continuous functions.

share|improve this answer
add comment

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.