I'm studying Topology for the first time, using Munkres(self study). Some time ago I first saw the definition of a continuous function:
\begin{equation} f:X\to Y \text{ is continuous iff for every open set }V\subset Y, f^{-1}(V) \text{ is open in X}. \end{equation}
I had no clue what this definition meant, until I saw this equivalent definition:
$f: X\to Y$ is continuous iff $\forall x\in X $ and for every open set $V$ containing $f(x)$, $\exists U\text{ s.t. }x\in U\text{ and }f(U) \subset V$, where $U$ is open.
which seems like a very natural generalization of continuity studied in analysis. So my first question is, why would you write the definition in the first form? Does this become natural after I study a lot of topology? It seems a little obscure to me.
Also, today I had a similar situation with the definition of the Quotient topology:
Definition. Let $X$ and $Y$ be topological spaces; let $p:X\to Y$ be a surjective map. The map $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.
This time I read it, and I read the paragraph after the definition which said about saturated sets. But this time, I really have absolutely no clue about this definition that I can't read more. I would like to know what this definition has to do with the quotient I saw in group theory or such.
It does become natural after a while, as it happens with most things that one works with for an extended period of time.
Regarding the definition itself, there is an excellent question on MathOverflow on a related topic.
One answer that I find satisfying says that, paraphrasing, all the concepts in general topology can be stated in terms of open sets, which then make defining a topology in terms of open sets an efficient approach. This is not to say one should always think of continuity, compactness, connectedness, etc. in terms of open sets.
I think the more intuitive way to think about this is to first define quotients of topological spaces. So let's give a motivation for that first, in analogy with their group theoretic counterpart.
Take a group $G$ and a normal subgroup $H$. The quotient group $G/H$ is defined as the equivalence classes of the relation $g \sim g' \iff gg^{-1} \in H$,
$$ G/H = \{gH : a \in G\} $$
toghether with the operation $gH \cdot g'H = gg'H$. Since $H$ is normal, this defines a group, and we have a canonical group morphism
$$ q : G \to G/H $$
that sends an element $g$ of $G$ to its class $gH$.
In general, given a group morphism $f : G \to G'$, its kernel is always a normal subgroup of $G$ and thus we can make sense of the quotient $G/\ker f$. Moreover, the first isomorphism theorem for groups says that
$$ G/\ker f \simeq \operatorname{im} f. $$
via
$$ \bar{f} : [g] \in G/\ker f \mapsto f(g) \in \operatorname{im} f \subset G' $$
Intuitively, the elements on the image are like the elements of $G$, up to multiplication by elements that are "trivial for $f$". When $f$ is surjective, we obtain
$$ G/\ker f \simeq G'. $$
Now, can we describe the equivalence relation previously defined in terms of $f$. Two elements $[g],[g']$ are equal in $G/\ker f$ precisely when $gg'^{-1} \in \ker f$, that is, when $1 = f(gg'^{-1}) = f(g)f(g')^{-1}$. In other terms,
$$ g \sim g' \iff f(g) = f(g'). $$
With this terminology, we know that for any group morphism $f : G \to G'$ we can define the equivalence relation $g \sim g' \iff f(g) = f(g')$, the equivalence classes have a group structure, and they are isomorphic to $\operatorname{im} f$.
Now let's go back to topological spaces. Take a space $X$ and an equivalence relation $R$ on $X$. We can define a topology on the equivalence classes $X/R$ by giving them the final topology with respect to the canonical projection $q : x \in X \mapsto [x] \in X/R$, that is, a set $U \subset X/R$ is open if and only if $q^{-1}(U)$ is open in $X$. One may argue that this is somewhat natural, as we want $q$ to be continuous and this topology is the minimal one that satisfies this.
Now, if we have a continuous $f : X \to Y$ that is compatible with $R$, i.e. such that $x \sim y$ implies $f(x) = f(y)$, we obtain a countinuous function
$$ \bar{f} : [x] \in X/R \mapsto f(x) \in Y $$
This is suspiciously similar to what we can do for groups. In effect, if $f : X \to Y$ is a continuous map, we can define an equivalence relation $R$ as
$$ x \sim y \iff f(x) = f(y) $$
exactly as we did we groups. Then $f$ is trivially compatible with $R$ and thus we obtain a map
$$ \bar{f} : [x] \in X/R \mapsto f(x) \in f(X) \subset Y, $$
again as in the group theoretic case. It would be nice to recover a version of the first isomorphism theorem in this setting, i.e. it would be nice for $\bar{f} : X \to f(X)$ to be a homeomorphism.
Since we are restricting ourselves to $f(X)$ anyway, we can assume $f$ surjective. It can be shown that $\bar{f}$ is bijective (the proof is identical to the one for groups), and we already know that $\bar{f}$ is continuous, so what is left to determine is whether $\bar{f}$ is open.
Take an open set $U \subset X/R$. By definition $U$ is open if and only if $q^{-1}(U)$ is open in $X$. Moreover, we know by surjectivity of $q$ that $U = qq^{-1}(U)$ and by defition of $\bar{f}$ we get $\bar{f}q =f$. Therefore our question reduces to analyzing when is
$$ \bar{f}(U) = \bar{f}(q(q^{-1}(U)) = f(q^{-1}(U)) $$ open. Once again by surjectivity, we have
$$ f^{-1}(fq^{-1}(U)) = q^{-1}(U) $$
which we assume open in $X$. Therefore, if $f$ were final, this proves that $\bar{f}(U)$ is open for every $U$ open in $X/R$. Reciprocally, if $\bar{f}$ were a homeomorphism, then
$$ V \text{ open in $Y$} \iff \bar{f}^{-1}(V) \text{ open in $X/R$} \iff q^{-1}\bar{f}^{-1}(V) = f^{-1}(V) \text{ open in $X$}. $$
We have finally obtained the following characterization,