Adjective
bijective- (of a map) both injective and surjective
- (of a set) having a bijective map
Read full definition at wiktionary.org
...Alternatively, f is bijective if it is a one-to-one correspondence between those sets; i.e., both one-to-one (injective) and onto (surjective). (One-to-one function means one-to-one correspondence (i.e., bijection) to some authors, but injection to others.) Read full entry
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- 1.Bijection - Wikipedia, the free encyclopedia
- In mathematics, a bijection, or a bijective function is a function f from a set ... Bijective functions play a fundamental role in many areas of mathematics, for ...
- http://en.wikipedia.org/wiki/B
ijective
- 2.Bijective proof - Wikipedia, the free encyclopedia
- "A direct bijective proof of the hook-length formula" – by Novelli, Pak and Stoyanovsky. "Bijective census and random generation of Eulerian planar maps with ...
- http://en.wikipedia.org/wiki/B
ijective_proof
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Can anyone give me an example I have to give an eg of a
continous one to one and on to
function between topological
spaces whose inverse is not
continious. Now from my
readings I figure this is an
example of a bijective
homomorphism? I could be wrong
(it happens often) but I need
to be able to give an example
with numbers or at least
something I can draw a sketch
of to prove it's inverse is
not continuous. Can anyone
help me or link me in the
right direction?
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A function is continuous if the inverse image of an open set is open. The bijective function to use is obvious - the identity map. The difference between the domain and range spaces is the topological structure. We require that the topology of the domain space is "thicker" than that of the range. So, a general map would be: f: (X,A) --> (X,B); f(x) = x, with the requirement that B is strictly contained in A. This will be continuous since the inverse image of an open set is itself and is open by the containment, and the strict containment will prevent continuity for the inverse. A specific example: f: (R,P(R)) --> (R, E), f(x) = x. (R, P(R)) is the reals with the discrete topology; (R, E) is the trivial topology with only R and the empty set open. Hope this helps. |
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Prove that a function has an Im doing a uni course on set
algebra and i missed the
lecture today. Im trying to
catch up, but i havent seen
any proofs of the like before.
Thanks.
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Note that this theorem assumes a definition of inverse that required it be defined on the entire codomain of f. Some books will only require inverses to be defined on the range of f, in which case a function only has to be injective to have an inverse. Obviously your current course assumes the former convention, but I mention it in case you ever take a course that uses the latter. Okay, to prove this theorem, we must show two things -- first that every bijective function has an inverse, and second that every function with an inverse is bijective. To prove the first, suppose that f:A → B is a bijection. Define the set g = {(y, x): (x, y)∈f}. I claim that g is a function from B to A, and that g = f⁻¹. First, we must prove g is a function from B to A. This means that we have to prove g is a relation from B to A, and that for every y in B, there exists a unique x in A such that (y, x)∈g. For the first part, note that if (y, x)∈g, then (x, y)∈f⊆A×B, so (y, x)∈B×A. This means g⊆B×A, so g is a relation from B to A. Next, let y∈g be arbitrary. Since f is surjective, there exists x such that f(x) = y -- i.e. (x, y)∈f, which means (y, x)∈g. Further, if z is any other element such that (y, z)∈g, then by the definition of g, (z, y)∈f -- i.e. f(z) = y = f(x), so z=x. Thus ∀y∈B, ∃!x∈A s.t. (y, x)∈g, so g:B → A is a function. Next, we must show that g = f⁻¹. Let x∈A be arbitrary. By the definition of function notation, (x, f(x))∈f, which by the definition of g means (f(x), x)∈g, which is to say g(f(x)) = x. Thus we have ∀x∈A, g(f(x))=x, so g∘f is the identity function on A. Similarly, let y∈B be arbitrary. Then (y, g(y))∈g, which by the definition of g implies that (g(y), y)∈f, so f(g(y)) = y. Thus ∀y∈B, f(g(y)) = y, so f∘g is the identity function on B. So g is indeed an inverse of f, and we are done with the first direction. To prove that invertible functions are bijective, suppose f:A → B has an inverse. Let x and y be any two elements of A, and suppose that f(x) = f(y). Then x = f⁻¹(f(x)) = f⁻¹(f(y)) = y. Therefore f is injective. To show that it is surjective, let x∈B be arbitrary. Then since f⁻¹ is defined on all of B, we can let y=f⁻¹(x), so f(y) = f(f⁻¹(x)) = x. Since we can find such y for any x∈B, it follows that if is also surjective, thus bijective. Q.E.D. |
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What's the difference btw a |
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A bijective function is both injective and surjective. If it's bijective then it's surjective the converse is not necessarily true (consider (x-1)(x-2)*x which is surjective but not bijective (since the roots all map to 0) ) A function f:A->B is surjective iff range(f)=B. f:R->R by f(x)=x^3 is surjective since it hits every real number but f(x)=x^2 is not since negative numbers dont have preimages. A function is injective iff f(x)=f(y) => x=y |
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