The following exercise can be found in Tao’s Analysis I.
Exercise 3.3.8(d) Show that if $X$ and $Y$ are disjoint sets and $f:X\to Y$ and $g:Y\to Z$ are functions, then there exists a unique function $h:X\cup Y\to Z$ such that $h\circ\iota_{X\to X\cup Y}=f$ and $h\circ\iota_{Y\to X\cup Y}=g$.
Now, here is my confusion: If $X\cap Y=\varnothing$, then clearly $\iota_{X\to X\cup Y}=\operatorname{id}_X$, as $\nexists x\in X : x\in Y$. Similarly, $\iota_{Y\to X\cup Y}=\text{id}_Y$. Substitution yields that for suitable $h$, $$h\circ \text{id}_X =f \ \ \land \ \ h\circ\text{id}_Y =g,$$ but this is meaningless — composition is not defined for these functions (as $X\neq Y\neq X\cup Y$). Granted, the original compositions were well-defined. Why have I run into this problem and moreover, how can I continue on the right track?
For sure there exists such $h$. Just define $$h(a)=\begin{cases} f(a), & \text{if $a\in X$,}\\ g(a), & \text{if $a\in Y$.}\\ \end{cases}$$
Then, since $\iota_{X\to X\cup Y}(x)=x$, obviously $h\circ \iota_{X\to X\cup Y}(a)=h(a)=f(a)$, for any $a\in X$. Analogous to $a\in Y$.
Now, suppose that $h'$ also satisfies $h'\circ \iota_{X\to X\cup Y}=f$ and $h'\circ \iota_{Y\to X\cup Y}=g$.
So, for any $a\in X$, $$ h'\circ \iota_{X\to X\cup Y}(a)=f(a)= h\circ \iota_{X\to X\cup Y}(a) $$ and by definition of $\iota_{X\to X\cup Y}$ we obtain $$ h'(a)=h(a), \quad \forall\ a \in X.$$
By same argument we get $$ h'(a)=h(a), \quad \forall\ a \in Y.$$
Then, $h=h'$ and there is only one such function.