Topological Definition of Continuity I am having some trouble understanding the mechanics of the topological definition of continuity.
It has been defined as follows:
A function $ f: \mathbb{R^n} \to \mathbb{R^m}$ is continuous if for every open set $V \subseteq \mathbb{R^m}$, the inverse image $f^{-1}(V) = \{x \in \mathbb{R^n} | f(x) \in V \subset \mathbb{R^m}\}$ is an open subset of $\mathbb{R^n}$.
Why, specifically, under the context of topology do we define continuty in terms of the inverse? Is there an equivalent definition of this that goes "the other way around"?
Furthermore, this is not an iff implication. That is, an open subset of $\mathbb{R^n}$ does not imply $V \subseteq \mathbb{R^m}$ is closed. Or am I incorrect?
 A: The topological notion of continuity (which is stated for any topological space - even not metric, not only the $\mathbb{R}^n$) is a generalisation of the intuitions you may have from the real analysis (with $\delta$s and $\epsilon$s). Think of a function $f:\mathbb{R} \to \mathbb{R}$.  If it is not continuous at some point you may choose the neighbourhood violating the definition.
Let's analyse a simple example: $\operatorname{signum} : \mathbb{R} \to \mathbb{R}$  - equal 1 for positive numbers and 0 otherwise. If we take an open set $X = (-1/2, 1/2)$, its counter image $\operatorname{signum}^{-1}(X) = (-\infty, 0]$ is not open (because the function is "broken" in 0).
The definition is "iff" in the sense that functions $f:\mathbb{R}^n \to \mathbb{R}^m$ are continuous in the $\delta - \epsilon$ sense iff they have this property. However, there are more interesting topologies than the one on $\mathbb{R}$, where the $\epsilon - \delta$ definition does not make sense, but this one allows us to define what does it mean for the function to be continuous.
I think drawing it could help you to get more intuitions if this is not enough ; )
A: Continuous functions have many important properties. One crucial one in $\mathbb{R}^n$ is that they preserve convergent sequences. In other words, if I have a sequence $x_{n}$ converging to $x$, then $f(x_{n})$ converges to $f(x)$. What does it mean for $f(x_{n})$ to converge to $f(x)$? From the topological perspective, it means that the sequence $f(x_{n})$ eventually enters (and stays in) every open set $V$ containing $f(x_{n})$. If $f^{-1}(V)$ is open, then since it contains $x$ the sequence $x_{n}$ must eventually enter and remain in it, since we have assumed $x_{n} \to x$. From this we can conclude that $f(x_{n})$ eventually enter and remain in $V$. 
The key point of this proof is that we are using information in the domain to prove stuff in the range/codomain, so we need to be able to "pull back" objects from the former to the latter. If $f^{-1}(V)$ is open for every open $V$, then all the open sets in the codomain ``come from" open sets in the domain. But if we insist instead that $f(V)$ is open for $V$ open, then the codomain may have many extra open sets that are not of the form $f(V)$, and we would be at a loss to prove anything about them.
A: There are several equivalent def'ns of continuity. Some of them are more suitable for certain problems than others. 
1.The inverse of an open set of the range is open in the domain.
2.The inverse of a closed set of the range is closed in the domain.


*For a given $p$ in the domain of $f:$ Suppose that whenever $f(p)\in V$ with $V$  open in the range, there is an open $U$ in the domain, with $p\in U,$ such that $\forall x'\in U\;(f(x')\in V).$ Then we say that  $f$ is continuous  at $p.$ And $f$ is continuous iff $f$ is continuous at every point $p$ of its domain. This is the topological generalization of the "$\delta, \epsilon$" criterion for continuity of a real function. 

*For a function $f:X\to Y$ : Whenever $S\subset X$ and $p\in Cl_X(S)$ then $$f(p)\in Cl_Y\{f(p'):p'\in S\}.$$ This is the topological generalization of a criterion for continuity of $f$ when $(X,d), (Y,e)$ are metric spaces: Whenever $(p_n)_n$ is a sequence in $X$ with $\lim_{n\to \infty}d(p,p_n)=0$ then $\lim_{n\to \infty}e(f(p),f(p_n))=0.$

*Local continuity: $f$ is continuous iff every $p$ in the domain of $f$ belongs to an open set $U_p$ in the domain such that the restriction of $f$ to the domain $U_p$ is continuous. This is useful in some constructions. E,g. where we have $X=\cup_{n\in N}U_n$ where each $U_n$ is open in $X, $ and $U_n\subset U_{n+1},$ and we have continuous $f_n:U_n\to Y$ with $f_{n+1}|U_n=f_n.$ Then $\cup_{n\in N}f_n$ is continuous.
A: Let $X, Y$ be two topological spaces.  A function $f : X \rightarrow Y$ is said to be continuous at a point $x_{0} \in X$ if into every neighborhood of $y_{0} = f(x_{0}) \in Y$ this function maps some neighborhood of $x_{0}$.
(See also the definitions of continuity and equivalent conditions in: (a) Komogorov and Fomin's book on real analysis, (b) Vulikh's Brief Course of Functions of a Real Variable.)
