TheoremBase

Gluing Maps on Disjoint Domains, and Shrinking the Codomain

Two maps on disjoint domains glue to a map on the union, and two bijections with disjoint domains and disjoint codomains glue to a bijection; a map may be regarded as a map into any class containing its image, and an injective map is a bijection onto its image.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let AA and CC be sets and XX a class, with maps and their values as in Functions, Values of a Function, and Functions from One Class to Another, and the image f(A)f(A) of AA under a map ff.

Gluing. Let f:A→Xf:A\to X and g:C→Xg:C\to X be maps with A∩C=∅A\cap C=\emptyset. Then f∪gf\cup g is a map f∪g:A∪C→Xf\cup g:A\cup C\to X, with (f∪g)(x)=f(x)(f\cup g)(x)=f(x) for x∈Ax\in A and (f∪g)(x)=g(x)(f\cup g)(x)=g(x) for x∈Cx\in C.

Gluing bijections. Let BB and DD be sets with B∩D=∅B\cap D=\emptyset, and let f:A→Bf:A\to B and g:C→Dg:C\to D be bijections with A∩C=∅A\cap C=\emptyset. Then f∪gf\cup g is a bijection f∪g:A∪C→B∪Df\cup g:A\cup C\to B\cup D.

Shrinking the codomain. Let f:A→Xf:A\to X be a map and YY a class with f(A)⊆Yf(A)\subseteq Y. Then f:A→Yf:A\to Y is a map. If ff is injective, then f:A→f(A)f:A\to f(A) is a bijection.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…