From pg. 64 of Categories for the Working Mathematician:
If $$ E = \{(fx, gx) \mid x \in X \} \subset Y \times Y, $$
I can see how $[\text{im}(f) \cup \text{im}(g)] / E$ makes sense, but might not $Y / E$ fail to make sense since there might be elements in $Y$ that aren't related by $E$?
On
Well let us look at an example. Let $X = \{1,2,3\}$ and $Y = \{1,2,3,4,5\}$ and let $f(x) = 2x - 1$ and $g(x) = x$. Then $E$ is the smallest relation on $Y \times Y$ containing the relations $1E1$, $3E2$,and $5E3$. Since $E$ needs to be reflexive, we add in $xEx$ for all $x \in Y$. Since $E$ needs to be symmetric, we add it $2E3$ and $3E5$. Since $E$ needs to be transitive, we add in $2E5$ (since $2E3E5$) and its reverse of course.
Thus the equivalence classes of $E$ are $[1] = \{1\}$, $[2] = \{2,3,5\}$ and $[4] = \{4\}$.
In an equivalence relation $E$ on a set $Y$, all $(y,y)$ are elements of $E$. So to construct the least equivalence relation containing some relation $R$ we must (i) throw in all $(y,y)$ for $y\in Y$, (ii) all $(x,y)$ and $(y,x)$ where $(x,y)\in R$ and (iii) all $(x,z)$ when we already have $(x,y)$ and $(y,z)\in R$.