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$f:X\rightarrow Y$ 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.

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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.
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could you check once more about $f$? – Un Chien Andalou 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. – Un Chien Andalou Jul 12 '12 at 11:27
@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

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

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Let $X=(0,1)\cup (2,3)$ in usual subspace topology, then locally constant functions are surjective? what about $1,2$ – Un Chien Andalou 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! – Host-website-on-iPage Jul 12 '12 at 11:15
okay got it :)) – Un Chien Andalou Jul 12 '12 at 11:17

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

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thank you :)))))))))) – Un Chien Andalou Jul 12 '12 at 11:17

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.

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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 $\Bbb{R}/\sim$.

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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.

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