$f:X\rightarrow Y$ is a continuous surjection, $X$ and $Y$ are topological space $f:X\rightarrow Y$ is a continuous surjection Then


*

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

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

*If $A$ is an infinite subset, does this imply that $f(A)$ is so in $Y$?
Thank you for help.
 A: 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.

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


*

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

*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 $Y$.

*$X=Y\coprod Y$ with $Y$ infinite and $f$ the identity on one component and constant on the other one. 
A: $3$ is false as you can construct functions that are locally constant and choose $X$ disconnected (each connected component infinite) for a counterexample.
A: 1 and 2 are false. Here are counterexamples for them:


*

*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.

*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 $\Bbb{R}/\sim$.
A: 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.

A: The first statement is true if $f$ is an open map.
The second statement is true if $f$ is a closed map.
