The sum of two sets and the disjoint union. I have the following question:


 

For(a) but I do not know how to show practically (using the givens) that $in_{S}$ must be injective, could anyone help me in doing so?
For(b),(c) I do not know how to prove them , could anyone help me in doing so please? 
Also, I do not understand what is the importance of $in_{S}$ and $in_{T}$ being injective in defining the disjoint union, could anyone explain this for me please?  
 A: This exercise is all about finding tricky $U, f, g$ and using the "fundamental" property (I guess a fundamental property is almost a universal property?), eg :


*

*a) $in_{S}$ is injective
As the only map $\emptyset \mapsto X$ is injective, WLOG $S\neq \emptyset$.
We use the following (easy) claim : if $v \circ f$ is injective, then $f$ is injective.
Take $U = S$, $f = \operatorname{Id}_S$, and $g : T \mapsto S$ any map (there is such a map since $S\neq \emptyset$).
We have $(f, g) \circ in_S = \operatorname{Id}_S$, which is injective. Hence, $in_S$ is injective.

*b) $\operatorname{Im}(in_S) \cap \operatorname{Im}(in_T) = \emptyset$
Assume for contradiction that there is $x \in \operatorname{Im}(in_S) \cap \operatorname{Im}(in_T)$.
Consider $U := \{0, 1\}$, $f$ the constant $0$ map $S \mapsto U$, and $g$ the constant $1$ map $S \mapsto U$.
Since $(f, g) \circ in_S = f$ we must have $(f, g)(x) = 0$.
But since $(f, g) \circ in_T = g$ we must have $(f, g)(x) = 1$. Contradiction! 

*c) $\operatorname{Im}(in_S) \cup \operatorname{Im}(in_T) = \operatorname{Im}(in_S) \sqcup \operatorname{Im}(in_T)$
This one is up to you, be creative (and look at @Greg Martin's excellent comment!)  :)

