# $X,Y$ topological spaces, $f:A\rightarrow Y$ continuous, $Y\rightarrow X\cup Y\rightarrow X\cup_f Y$ embedding

Let $X,Y$ both be topological spaces, $A\subseteq X$ and $f:A\rightarrow Y$ a continuous function. Take the disjoint union of $X$ and $Y$, $X\cup Y$, and then identify each $a \in A \subseteq X$ with $f(a) \in Y$ and take the quotient topology on the space and denote this space by $X \cup_f Y$. Show that the composition of the inclusion map with the quotient map, $$Y\rightarrow X \cup Y\rightarrow X\cup_f Y$$ is an embedding.

This was an exam question. I am wondering what the space $X \cup_f Y$ looks like and what is the quotient map from $X\cup Y\rightarrow X\cup_f Y$. I have looked at other Topology textbooks, we're using Munkres' topology second edition, and I've seen that quotient spaces are created from equivalence classes. What I am most familiar with in terms of quotient spaces is if we have a map $p:X\rightarrow Y$ which is surjective, and $X$ is a topological space, then we can endow $Y$ with the topology where $$U \text{ is open in Y if and only if } p^{-1}(U) \text{ is open in X,}$$ and we would call $p$ a quotient map and $Y$ would be called the quotient space. Feel free to correct me on terminology. That is what I am most familiar with, a quotient map, not too much creating a quotient space via an equivalence relation. Could someone quantify the equivalence relation which gives rise to $X\cup_f Y$ and what set the equivalence relation is on and what is the map $$X\cup Y\rightarrow X\cup_f Y$$ which we call the quotient map? I am guessing it (hopefully) won't be difficult to show that the map is an embedding once I know what the space $X\cup_f Y$ exactly is and then what is the map $X\cup Y\rightarrow X\cup_f Y$. Let me know if more details are needed.

• $$X\cup_f Y = {\rm graph}(f) \cup X\setminus A$$ Or in other words, it looks like $X$ but all values $x$ "glued" together whenever $f(x) = f(x')$... For $X = Y = \mathbb R$ and $f: x\mapsto x^2$ you have $$X\cup_f Y = \{[x] : x \in \mathbb R_{\ge 0}\}$$ – AlexR Nov 5 '13 at 21:19

Let's keep your definition of a quotient space as a space $Y$ with a map $q:X\to Y$ such that $U\subseteq Y$ is open iff $q^{-1}[U]$ is open in $X$. Let's call an identification space a space $X/\sim$ where $\sim$ is an equivalence relation on $X$, that means $X/\sim$ is the set of equivalence classes $[x]$ of points $x$ in $X$. Since we want to equip this set with a topology, we ask ourself: What could make a natural candidate for an open set? Any set $U$ in $X/∼$ is a set of equivalence classes. This is a rather abstract set (a set whose elements are sets), so let's look at a simple example:
If $X=[0,1]=I$ and the equivalence relation is given by $x∼y$ if $x=y$ or $x,y\in\{0,1\}$, then only one class is non-trivial, namely $[0]=[1]=\{0,1\}$, all other classes are singletons $[x]=\{x\}, x\ne0,1$. So we could think of $X/∼$ as $I$ itself with two points $0,1$ replaced by the class $\{0,1\}$, that means $0$ and $1$ become equal in $X/∼$. Intuitively, we are gluing $0$ and $1$ together, so the points near one end of the interval become close to the point near the other end of $I$. One can imagine $I/∼$ as a string with their ends tied together. Since this gives us kind of a loop where a neighborhood of $0=1$ should look like an arc around this point, we see that this set is a neighborhood around both $0$ and $1$ if we untie the knot.
So a set $U$ in $X/∼$ is regarded open if its preimage is open in $X$. Another characterization is:
The open sets in $X/∼$ are precisely the images of open saturated sets in $X$.
A set is saturated if $U=f^{-1}[f[U]]$
So we see that an identification space is a quotient space. On the other hand, a quotient space $Y$ can be considered an identification space, namely $Y\approx X/∼$, where $x∼y$ iff $q(x)=q(y)$. That's why the term are used interchangeably.

The map $X\sqcup Y\to X\cup_fY$ is the quotient map sending a point in $x\in X\setminus A⊔Y\setminus f[A]$ to $\{x\}$, and $a\in A$ as well as $f(a)\in Y$ to the class $\{f(a),b\mid f(b)=f(a)\}$. Often when dealing with quotient spaces, we write just "$x$" instead of "$[x]$", and we regard a subset $B$ as a subset of the quotient space if no identifications take place on $B$. In your case, we then consider $Y$ a subset of $X∪_fY$.
Now this doesn't imply that $Y$ is a subspace of $X∪_fY$. It would be if $q|_Y:Y→q[Y]$ were a quotient map, but unfortunately the restriction of quotient maps need not be a quotient map. However, here it is, and to prove it we have to show that an open set $U\subseteq Y$ has an open image in $Y$, i.e. $q[U]$ is the intersection of an open $V⊆X∪_fY$ with $Y$. This again means that $q^{-1}[V]$ must be open in $X⊔Y$, i.e. open in $Y$ and open in $X$.
I'll leave it to you to try and find such a set. Hint: The continuity of $f$ is important.

• Could I ask what is the equivalence relation on $X\coprod Y$ that gives rise to the classes $\{x\}$ for $x\in X\setminus A\cup Y\setminus f(A)$ and the other class being $\{f(a),b\mid f(b)=f(a)\}$? I see the example with $I=[0,1]$ and $x\sim y$ if and only if $x=y$ or $x,y \in \{0,1\}$. Thank you for the explanation it is enlightening. We didn't cover this very much in the course. – Frudrururu Nov 19 '13 at 18:22
• We identify $a∈A$ with $f(a)∈Y$. Of course this is not an equivalence relation, it is only symmetric (because we simultaneously mean $a∼f(a)$ and $f(a)∼a$). So we add the reflexivity by $x∼x$ for all points, and then we take the transitive closure, that means we identify $v∼w$ whenever there is a finite sequence $v=v_1,v_2,...v_n=w$ such that $v_i∼v_{i+1}$. In your case it amounts to $a∼b$ whenever $f(a)=f(b)$. Note that here it is possible to explicitly mention each identification, but sometimes there are "too many", then we just describe it as the equivalence relation generated by $a∼f(a)$ – Stefan Hamcke Nov 19 '13 at 18:43

This space, $X\cup_f Y$, is called an adjunction space, it connects topological spaces along the values of a function $f:X\rightarrow Y$. Namely, for $a \in X$, $a \sim f(a)$ where $f(a) \in Y$. This is often seen when working with connected sums, see here. Basically you connect the spaces along this function. In the case of a connected sum, you connect them along some boundary by identifying the points. I hope this gives you enough detail to show it is an embedding, if you need more feel free to ask.