What is the Quotient on the Coproduct in Adjunction Spaces

Question

Can someone please provide a detailed explanation of the equivalence relation used to construct adjunction spaces from the topological coproduct? In particular, most sources talk about "identifying an element with its image", which generally live in different spaces, but equivalence relations are defined on the cartesian product of two copies of some set. So what exactly is the relation here and why can one quotient by it (why is it an equivalence relation)?

Thanks in advance.

Answer 1

The adjunction space $X\cup_f Y$ for a function $f\colon A\to Y$ with $A$ a subspace of $X$ is given by $X\cup_f Y = (X\sqcup Y)/{\sim}$ where $\sqcup$ is the disjoint union and $\sim$ is the equivalence relation given by $$\begin{array}{rc} z\sim z & \forall z\in X\sqcup Y \\ a\sim f(a), f(a)\sim a & \forall a\in A\\ a\sim a' & \forall a,a'\in A\mbox{ s.t. } f(a)=f(a')\end{array}$$ which can be shown to be an equivalence relation. As was said in another answer, this is the same relation as the one generated by $a\sim f(a),\:\: \forall a\in A$.

Note that the setup and notation that I have used is slightly different to the other answer. They are equivalent, although mine is less symmetrical because I've explicitly stated that $A$ should be a subspace of $X$ and then there is only one map from $A$ to $Y$. If we have some $Z$ and map $g\colon Z\to X$ and $h\colon Z\to Y$. For convenience, let's assume that $g$ is injective - if not, we can still form an equivalence but it is more work. Then if we let $A=g(Z)$ and define $f\colon A\to Y$ by $f(a)=h(g^{-1}(a))$ then we see that $X\cup_f Y = X\bigsqcup_Z Y$.

Answer 2

If you have two continuous maps $f:Z\rightarrow X$ and $g:Z\rightarrow Y$, the adjunction space (or pushout, as it is usually called), is the space $W=X\coprod_ZY=(X\coprod Y)/\sim$, where $\sim$ is the smallest equivalence relation on $X\coprod Y$ which contains the elements $f(z)\sim g(z)$ for every $z\in Z$. This notion of an equivalence relation on a set $A$ generated by a set $S\subset A\times A$ is fundamental here. It is the smallest equivalence relation $R$ on $A$ such that $S\subseteq R$. This exists by Zorn's lemma (see this question). Intuitively, $R$ is "constructed" by throwing in the minimal number of elements necessary to make the axioms of an equivalence relation hold.

Answer 3

I like to see this as that you form the set $Q=(X \setminus A) \sqcup Y$ and the function $p: X \sqcup Y \to Q$ which is the identity on $X \setminus A$ and on $Y$ and sends points $a \in A$ to $f(a)$. Then $Q$ with the identification (quotient) topology is the adjunction space you want.