Understanding an Example of a *Pushout* in $\mathbf{Set}$ 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 $x \in a$?
 A: This indeed is confusing. First of all, the pushout is not the disjoint union, it is the disjoint union with the images of $a$ identified.
Here's a specific example, designed to hit some "conceptual edge cases." Suppose
$$a = \{1,2\}$$
$$b = \{0,1,2,3\}$$
$$c = \{0,1,2,3,4\}$$
and for $f:a\to b$ and $g:a \to c$,
$$f(x) = x+1$$
$$g(x) = x.$$
Let's call the terminal $r$ as in the first picture.
We'll keep elements separate by adding little subscripts for whether they came from set $b$ or set $c$.
Thus, as described, $r$ contains $0_b,1_b,2_b,3_b$ and $0_c,1_c,2_c,3_c,4_c$, but these are not all distinct.
In particular, we must identify $f(x)$ and $g(x)$ for each $x\in a$.
This means in $r$ we set $f(1)=2_b=1_c=g(1)$ and $f(2)=3_b=2_c=g(2)$.
So the elements in $r$ are:
$$\{ 0_b,$$
$$0_c,$$
$$1_b,$$
$$2_b=1_c,$$
$$3_b=2_c,$$
$$3_c,$$
$$4_c\}$$
A: 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.
A: The ‘disjoint union’ of $b$ and $c$ does not mean the sets $b$ and $c$ are necessarily disjoint: it means copies of $b$ and $c$ are made so theses copies are disjoint, viz. $b_1=\{(x,1)\mid x\in b\}$ and $c_2=\{(x,2)\mid x\in c\}$. It is the coproduct in the category of sets (and also in the category of topological spaces).
