TITLE: Understanding an Example of a *Pushout* in $\mathbf{Set}$
QUESTION [7 upvotes]: On pg. 66 of Categories for the Working Mathematician, the notion of the pushout is defined, along with the notion of a "fibered sum":



The author then claims:

In $\mathbf{Set}$, the pushout of $\langle f, g \rangle$ always exists; it is the disjoint union of $b \amalg c$ with the elements of $fx$ and $gx$ identified for each $x \in a$.

Can someone clarify what is meant by this last example in $\mathbf{Set}$? How do we know that $b$ and $c$ form a disjoint union? Couldn't $fx = gx$ for some $x \in a$?

REPLY [5 votes]: The operations of 'union' and 'disjoint union' are different. The union $A \cup B$ of sets $A$ and $B$ is the set of all objects which are elements of $A$ or elements of $B$. The disjoint union $A \sqcup B$, on the other hand, builds in some kind of identifier that allows you to tell whether the element came from $A$ or from $B$. A common definition is
$$A \sqcup B = \{ (a,0) \mid a \in A \} \cup \{ (b,1) \mid b \in B \}$$
There is always a surjection $A \sqcup B \to A \cup B$ given by $(x,i) \mapsto x$; in the event that $A$ and $B$ are disjoint, this surjection is in fact a bijection.
Thus, the pushout of $f : X \to Y$ and $g : X \to Z$ can be taken to be the quotient of $Y \sqcup Z$ by the least equivalence relation identifying $(f(x),0)$ and $(g(x),1)$ for all $x \in X$.
In fact, it's funny that you should mention pushouts, since the union of sets can be constructed as a pushout. Given sets $A$ and $B$, we have $A \cap B \subseteq A$ and $A \cap B \subseteq B$. Let $i_A : A \cap B \to A$ and $i_B : A \cap B \to B$ be the inclusions. Then $A \cup B$ is (naturally isomorphic to) the pushout of $i_A$ and $i_B$. Working through the verification of this fact would be a good exercise to help you understand pushouts.