The sum of two sets and the disjoint union.

Question

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?

Answer 1

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!) :)