Show that $f$ is continuous between topological spaces if and only if $f$ is continuous at every $x \in X$. A mapping $f$ between topological spaces $(X, \mathcal I_X)$ and $(Y, \mathcal I_Y)$ is continuous in $x \in X$ if $f^{-1}(V)$ is a neighborhood of $x$ for every neighborhood of $f(x)$.
However $f$ is continuous between topological spaces if $f^{-1}(V) \in \mathcal I_X$ for every $V \in \mathcal I_Y$.
I want to show that $f$ is continuous between topological spaces $\iff$ $f$ is continuous at every $x \in X$.
I've already proved the direction $\Rightarrow$.
The other direction $\Leftarrow$ is given me headache. I've tried to consider $V \in \mathcal I_Y$, but cannot find a proper way to apply the assumption made.
Can someone help me out ?
 A: Let $f$ be continuous at everu $x$. 
$V \in I_Y$ be an open set. 
You have to prove that $f^{-1}(V)$ is an open set. If you succeed in proving this, $f$ will be "topological continuous".
Suppose that $V$ is nonempty (otherwise it's trivial) and let $y$ be in $V$. Then, $V$ is a neighborhood of $y$ in $Y$. 
If $f^{-1}(V) = \emptyset$, there is nothing to say : $\emptyset$ is open. 
If not, then write $U = f^{-1}(V)$ and then pick $x \in U$ ; let $y$ be $f(x)$. Then, $y$ is in $V$. Let's prove that $U$ is a neghborhood of $x$. 
You supposed that $f$ is continuous at each point of  therefore it is continuous at $x$. Take a neighborhood $V_y$ of $y$, contained in $V$. Then, $f^{-1}(V_y)$ is a neighborhood of $x$, obviously contained in $U = f^{-1}(V)$.
Therefore, $U = f^{-1}(V)$ contains a neighborhood of each of its points : it is an open set. This proves that $f$ is topological continuous.
A: Hint: A set $O$ is open $\iff$ $\forall x \in O: \exists$ open $ U$ s.t. $x \in U \subseteq O$. Use this to show that $f^{-1}(V)$ is open. 
A: For every $x\in f^{-1}(V)$ we have $f(x)=y\in V$. So there is a neighborhood $U_x\subset V$ of $y=f(x)$ with $f^{-1}(U_x)\subset f^{-1}(V)$. Note finally $ f^{-1}(V)= \bigcup_{x\in f^{-1}(V)}f^{-1}(U_x)$.
A: Let $f:X \rightarrow Y$ a function then for all $U \subseteq X$ one has $U \subseteq f^{-1}(f(U))$ (to be used soon).
Now let $V \in I_Y$, then for all $x \in f^{-1}(V)$ one has $f(x) \in V$. 
$f$ is continuous at $x$ so one has an open set $U_x$ s.t $x \in U_x$ and $f(U_x) \subseteq V$. 
Taking $f^{-1}$ on this result one has: $$U_x \subseteq f^{-1}(f(U_x)) \subseteq f^{-1}(V)$$
So one has: $$\cup_{x \in f^{-1}(V)}U_x \subseteq f^{-1}(V)$$ 
And obviously: $$f^{-1}(V) \subseteq \cup_{x \in f^{-1}(V)}U_x$$ 
Unifying all the $U_x$'s, one has a union of open sets which is open, hence is in $I_X$ so one has:
$$f^{-1}(V) = \cup_{x \in f^{-1}(V)}U_x \in I_X$$
