Feb 1, 2026 · If the natural numbers are bijective onto themselves, how can they also be bijective onto the integers? Ask Question Asked 23 days ago Modified 4 days ago

To prove a function is bijective, you need to prove that it is injective and also surjective. "Injective" means no two elements in the domain of the function gets mapped to the same image.

Nov 22, 2021 · What's the difference between Injective and Bijective? For example, is there a more rigorous proof of the bijectivity of a function? Also, can these properties be applied to more than just …

Update: In the category of sets, an epimorphism is a surjective map and a monomorphism is an injective map. As is mentioned in the morphisms question, the usual notation is $\\rightarrowtail$ or $\\

Sep 3, 2017 · I know that in order for a function to be invertible, it must be bijective, but does that mean that all bijective functions are invertible?

Is the following true: $g\\circ f$ bijective iff $f$ and $g$ bijective? Or can the requirements be weakened for $g$ (i.e. $g$ only injective or surjective)? Or $f$?

$ (g \circ f)$ is bijective $\rightarrow$ $f$ is bijective, $f: X \rightarrow Y$ $\hspace {.5cm} g:Y\rightarrow Z$ All of the domains and codomains here are supposed to be the real numbers.

Nov 6, 2015 · In stead of this I would recommend to prove the more structural statement: "f:A→B f: A → B is a bijection if and only if it has an inverse". An inverse is a map g:B→A g: B → A that satisfies …

Jun 20, 2019 · In categories of algebras yes, a morphism that is bijective on underlying sets is always an isomorphism (invertible morphism). You want to restrict to algebras-and-operations, as relational …

But anyways. Objective: Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism Proof: (Honestly not sure what I am doing but …