# Disjoint union

2010 Mathematics Subject Classification: Primary: 03E [MSN][ZBL]

discriminated union, sum

A construction in set theory corresponding to the coproduct, the union of disjoint "copies" of sets in a family. Let $X_\lambda$ be a family of sets indexed by $\lambda \in \Lambda$. The disjoint union $Y = \coprod_{\lambda \in \Lambda} X_\lambda$ has a universal property: there are maps $i_\lambda : X_\lambda \rightarrow Y$ such that for any family of maps $f_\lambda : X_\lambda \rightarrow Z$ for some $Z$ and all $\lambda \in \Lambda$, there is a map $F : \coprod_{\lambda \in \Lambda} X_\lambda \rightarrow Z$ such that $i_\lambda \circ F = f_\lambda$.

If the $X_\lambda$ are mutually disjoint, so that $\lambda \neq \mu \Rightarrow X_\lambda \cap X_\mu = \emptyset$, then their union $Y = \bigcup_{\lambda \in \Lambda} X_\lambda$ is said to be the (internal) disjoint union of the $X_\lambda$: one also says that the $X_\lambda$ form a partition or decomposition of $Y$. The $i_\lambda$ are the inclusion maps of the $X_\lambda$ into $Y$.

More generally, we may construct a disjoint union given any family $X_\lambda$ as follows. Let $Y' = \bigcup_{\lambda \in \Lambda} X_\lambda$ and define maps $i_\lambda : X_\lambda \rightarrow Y' \times \Lambda$ by $i_\lambda : x \mapsto (x,\lambda)$. Then each $i_\lambda$ is an injection, the images of the $i_\lambda$ are disjoint, and $Y = \bigcup_{\lambda \in \Lambda} \mathrm{im}(i_\lambda)$ is the (external) disjoint union of the $X_\lambda$.

A bouquet or wedge is a disjoint union of pointed sets. It has the same universal property with respective to pointed maps. There is a similar explicit construction.

How to Cite This Entry:
Disjoint union. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Disjoint_union&oldid=35410