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From my limited exposure to the term "natural equivalence", I form the idea that "natural equivalence" is a fanciful way of saying "bijection" (?!).

For instance, in the above text (from Switzer's Algebraic Topology Book), by looking at the proof, it seems that the proof of natural equivalence is just proving one-one correspondence (i.e. bijection).

I am quite sure that "natural equivalence"="bijection" is not correct, so how do I interpret it correctly?

Thanks!

yoyostein
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    The choice of the word equivalence is poor in my opinion. In general, a natural something is a natural transformation that is something in each component. So here natural bijection or natural isomorphism would have been a better choice. Technically, the authors are not wrong as sets are (discrete) categories and equivalences between such reflect as bijections, but stil... – Pece Jan 13 '17 at 16:05

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It is possible that indeed the authors are using the term natural just to say that the bijection is the obvious one, meaning that they are using the word natural in a not-categorical way.

On the other hand that bijection is actually natural in the sense of category theory: that is the mappings $(i_X \colon [*,X] \to \pi_0(X))_X$ make commute the diagrams $$\require{AMScd} \begin{CD} [*,X] @>{i_X}>> \pi_0(X) \\ @V[*,f]VV @VV\pi_0(f)V \\ [*,Y] @>>{i_Y}> \pi_0(Y) \end{CD}$$ where $f \colon X \to Y$ is a continuous map between topological spaces, $[*,f]$ and $\pi_0(f)$ are the images of $f$ through the functors $[*,-],\pi_0 \colon \mathbf{Top} \to \mathbf{Set}$.

The proof that the $i_X$'s form a natural equivalence it is a matter of simple calculations (i.e. verifying the commutativity of the diagrams as above).

Giorgio Mossa
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    Perhaps it is useful to give an example of an equivalence that is not natural. The typical example is that any finite-dimensional vector space is isomorphic to its dual, but this correspondence is not natural. See for example this question: https://math.stackexchange.com/questions/622589/in-categorical-terms-why-is-there-no-canonical-isomorphism-from-a-finite-dimens – Mathematician 42 Jan 13 '17 at 13:21
  • One criterion for "non natural" is that choices are involved, as in the isomorphism between a finite dimensional vector space over a field $k$ and the specific vector space $k^n$. But the full definition of "natural" involves functors and so categories. – Ronnie Brown Jan 13 '17 at 15:09