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What is a natural bijection?

  1. Aug 19, 2010 #1
    Seems like a silly question, but a search of the forum and Google and my online textbook yielded no results (*shakes fist at textbook writer). Please help?
  2. jcsd
  3. Aug 19, 2010 #2
    The question is best answered by category theory: a natural bijection is a natural isomorphism in the category of sets. (Since you're asking this question, I'm assuming you know nothing about category theory.) Conceptually, a "natural" map is one that may be defined independent of the particular object you're considering.

    For example: given two sets A and B, there is a bijection ηA,B: A × B → B × A, defined by ηA,B(a, b) = (b, a). Notice that this definition doesn't depend on the particular sets A and B. In particular, given any f: A → A' and g: B → B', they give a function f × g: A × B → A' × B' taking (a, b) to (f(a), g(b)), and similarly a function (g × f): B × A → B' × A. You can check that
    ηA',B' ∘ (f × g) = (g × f) ∘ ηA,B.​
    This equation is actually what it means for η to be a natural transformation, so we call η a natural bijection. (In category theory terms, the maps F: ((A, B) ⟼ A × B; (f, g) ⟼ f × g) and G: ((A, B) ⟼ B × A; (f, g) ⟼ g × f) are functors from the category Set × Set to Set, and η is a natural isomorphism from F to G.)

    Another classic example, not of sets this time, but involves linear algebra: If V is a real vector space, let V* = {f: V → R | f is linear} be the dual space of V. Then V* is a vector space with the vector operations defined pointwise. If T: V → W is a linear map, then define the linear map T*: W* → V* by T*(f)(v) = f(T(v)) (more concisely: T*(f) = f ∘ T).

    If V is finite-dimensional, then it has a basis {e1, ... en}; define a linear map φV: V → V* by φV(ei)(ej) = δij. This map is actually a bijection (isomorphism), but it is not natural: it depends on the choice of basis of V. (In category theory terms: the identity functor and the dual space functor are not naturally isomorphic; in fact, the identity functor is covariant and the dual space functor is contravariant, making such a thing impossible.)

    However: there is an isomorphism ηV from V to the double dual space V** = (V*)*, defined by ηV(x)(f) = f(x). Note that this definition doesn't depend on the choice of basis (or generally on the particular structure of V), and in fact, the following equation holds for any linear map T: V → W:
    ηW ∘ T = T** ∘ ηV.​
    Therefore η is a natural isomorphism. (Category-theoretically: η is a natural isomorphism from the identity functor to the double dual functor (which takes V to V** and T: V → W to T**: V** → W**) in the category of finite-dimensional real vector spaces.)

    It really is impossible to give a general definition of naturality without talking about category theory, so if you're really interested you should read about that. For a terse Wikipedia guide to natural transformations, read these:
    1. Category (mathematics)
    2. Functor
    3. Natural transformation
    (Wikipedia is, as usual, not the best way to learn about a new topic, but there are plenty of resources out there.)
    Last edited: Aug 20, 2010
  4. Aug 20, 2010 #3


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    Basically, it is an bijection that you can construct "without making unatural choices". The classic example is that if V,W are two vector spaces of the same dimension, then given any choice of a basis of V and of W, the linear map that sends one basis to the other is an isomorphism. But this isomorphism is not natural, because there is no natural choice of basis for V and W. Every basis is on the same "footing" and there is no reason to prefer one over the other. In other words, there is no reason to prefer one isomorphism over the other: there is none that feels more natural than the others.

    However, given any vector space V, then in the case W=V**, there is an isomorphism that feels natural. It is the isomorphism η that take x in V and sends it to the linear map V*-->R η(x)(f):=f(x). (See the above post) Again, I say it feels natural because there is no awkward choices involved in the definition of the isomorphism.

    If you study algebraic topology, you will encounter tons of examples of natural and unnatural maps.

    The category-theory definition gives a rigorous meaning to "natural", but I don't know if it is of any use besides fixing the vocabulary. In any case, when you start to get a feel for what natural means, you will rarely need to check with the category-theory definition to determine whether a map is natural or not, because it is first a notion that appeals to the intuition.
  5. Aug 20, 2010 #4


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    It is way more than just terminology. Natural isomorphisms/transformations are all around the place in category theory, notably with (co)limits and adjoint functors. There are a lot of useful results, like "every functor which has a left-adjoint commutes with limits", where natural isomorphisms and the like are relevant.
  6. Aug 20, 2010 #5
    Thanks guys, really appreciate the help. Speaking of resources, any recommendations for material readily available online? I'm just going looking through some abstract and linear algebra on my own time. Currently I'm working through a free ebook at http://www.math.miami.edu/~ec/book/book.pdf" [Broken], but I'd assume that there are better ones out there.
    Last edited by a moderator: May 4, 2017
  7. Aug 20, 2010 #6


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    Ah, cool!
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