Point-Set Topology

* The note was originally written in Korean, and later translated into English.

I. Quotient Topology and Equivalence Class

 Quotient topology is a crucial concept closely related to the fundamental group in algebraic topology. Before considering the quotient in a topological space, it's helpful to review how algebraic structures were formed using quotients in algebra.
 In algebra, performing a quotient on a Group or a Ring involved creating an equivalence relation based on a normal subgroup or an ideal, and then forming a smaller group or ring using cosets. This was analogous to constructing a new algebraic structure from the fibers (equivalence classes where morphism results are equal) of an equivalence relation given by a homomorphism (Isomorphism Theorem).
 The concept of a quotient in topological spaces is similar. Since there's no algebraic structure, it's defined either as a quotient map, which is a type of continuous surjective map, or as the coarsest possible topology that ensures continuity. While Munkres defines the quotient map first and then uses it to define the quotient topology, this chapter will define it more intuitively, with Munkres's definition to be discussed later.
 A quotient topology is defined as the coarsest topology on $Y$ such that a surjective map $p: X \rightarrow Y$ is continuous. The desire for continuity stems from the need for the original topological structure to be well-transferred to the quotient space, mirroring the use of homomorphisms for quotients in algebra.
 What structure must be satisfied for continuity? We can define an open set $U$ in $Y$ such that $p^{-1}(U)$ is an open set in $X$. In other words, if the preimage of a set is open, it is included in $\mathcal{T}_Y$. Are there any other sets to consider besides these? Due to the properties of preimages, no; if such sets are collected, preimages preserve unions, intersections, the entire set, and the empty set, ultimately forming a topology. This topology becomes the quotient topology induced by the surjective map $p$. $$ \mathcal{T}_{X/p}\mathrm{(Quotient\ topology\ induced\ by\ p)}=\{ U \in X/p | p^{-1}(U) \in \mathcal{T}_X\} $$ Above, $Y$ was subtly replaced with $X/p$ to emphasize the quotient. Anyway, a map to a topological space induced by this definition is defined as a quotient map.
 Meanwhile, the very idea of considering the image $Y$ of $X$ as a new topological space is ultimately equivalent to considering it as a topological space composed of fibers. From this perspective, a continuous surjective map $p$ must map open sets formed by the union of these fibers to open sets. Munkres defines such sets, which can be composed of the union of fibers, as saturated sets. The specific wording is as follows:
 A saturated subset $C$ of $X$ for a surjective map $p$ is a set that contains every $p^{-1}(\{y\})$ that intersects $C$.
 Since every point in $X$ belongs to a fiber, this is equivalent to a set composed of the union of fibers. By the same logic, a saturated subset $C$ is ultimately equivalent to finding a set that is $p^{-1}(p(C))$ with respect to map $p$.
 Ultimately, the minimum condition for deriving a topological space from the equivalence classes of an equivalence relation with respect to a continuous surjective map is that saturated open sets must be mapped to open sets. Defining such a map as a 'quotient map' is also equivalent to the definition above.

II. Quotient Map and Quotient Topology

 Munkres first defines a 'quotient map' and then defines the topology induced by such a quotient map as a quotient topology. A quotient map is defined for two topological spaces $X$ and $Y$ as a surjective map $p: X \rightarrow Y$ such that a subset $U$ of $Y$ is open if and only if $p^{-1}(U)$ is open in $X$. When the quotient map is defined this way, the induced topology naturally becomes the same as the definition of quotient topology written above.
 On the other hand, as can be seen from the definition, a quotient map is always continuous by definition, but it is not necessarily homeomorphic. This is because there is no guarantee that every $U \in \mathcal{T}$ appears as a preimage. Therefore, if bijectivity is added, it naturally becomes a homeomorphism.