$f:V_1\to V_2$ continuous impies that $f$ pulls back closed sets to closed sets Let $f: V_1 \to V_2$ be a continuous function, $V_1$ and $V_2$ be normed vector spaces. Need to prove that $U\subseteq V_2$ is closed implies $f^{-1}(U)$ is closed.

Proof:

Since $U$ is a closet set, it contains all its limit points. Let $y\in U$, then $\exists \{y_n\}\subseteq U$ such that $\lim\limits_{n\to\infty}y_n=y$. Since $f$ is continuous, $\exists x_n\in f^{-1}(U)$ such that $f(x_n)=y_n$, and $\lim\limits_{n\to\infty} f(x_n)=f(x)=y\in U$. This implies that $\lim\limits_{n\to\infty} x_n = x = f^{-1}(y)\in f^{-1}(U)$. Thus $x$ is a limit point of $f^{-1}(U)$, and $f^{-1}(U)$ is closed, as required.
Please let me know if my proof is OK.
 A: Since comments are not for extended discussions, and the question asks about proof checking,
I wrote some of the comment I made as an answer.
There are several problems with the proof, and the statement:


*

*The statement "U is closed if and only if $f^{-1}(U)$ is closed" is actually false: 
consider the function $f$ from $\mathbb R$ into $\mathbb R$ defined by $f(x)=x^2$ and $U=(-1,\infty)$. Then $f^{-1}(U)=\mathbb R$ is closed but $U$ isn't.

*A true statement would be "if $U$ is closed, then $f^{-1}(U)$ is also closed" (the only if  part of the OP's statement). This is actually the title of the question, which is different from the statements in the body and in the comments.

*Another fix would be to assume that $f$ is an homeomorphism. 

*The statement does not mean that "the pullback of closed sets are mapped onto closed sets" which means "if $U$ is closed, then $f(f^{-1}(U))$ is also 
closed", since the pullback of $U$ is $f^{-1}(U)$. The statement in the body of the question means "$U$ is closed if and only if its pullback $f^{-1}(U)$ is closed".
Again, the statement in the title is the correct one.

*The proof is wrong in several places. First of all, as pointed out by learnmore's answer, a proof should start by considering $y$ in $f^{-1}(U)$.

*The fact that $y_n$ belongs to $U$ does not imply that there exists $x_n$ in $f^{-1}(U)$ such that $f(x_n)=y_n$. Indeed, consider the same function 
$f(x)=x^2$ and in the first example, but this time $U=[-1,\infty)$, which is closed. Then, for $y_n=-1\in U$, there is no $x_n$ satisfying $f(x_n)=y_n$. 
$x_n\in f^{-1}(U)$ only means that $f(x_n)\in U$. But $f(f^{-1}(U))$ is only a subset of $U$ in general, and actually, "for any subset $U$, $f(f^{-1}(U))=U$"
is equivalent to the surjectivity of $f$. 

*May the sequence $\{x_n\}$ 
exists, there is no reason it has a limit.

*The proof assumes incorrectly that the inverse function $f^{-1}$ exists, which is false in general. 

A: The proof is not okay..
In order to show that $f^{-1}(U)$ is closed start with a limit point say $x$ of $f^{-1}(U)$.
We can find a sequence $x_n\in f^{-1}(U)$ such that $x_n\to x$.
Now $x_n\in f^{-1}(U)\implies f(x_n)\in U$. As $f$ is continuous $x_n\to x\implies f(x_n)\to f(x)$ 
As $U$ is closed so it contains all its limit points hence $f(x_n)\in U;f(x_n)\to f(x)\implies f(x)\in U\implies x\in  f^{-1}(U)$
