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Source: https://en.wikipedia.org/wiki/Seifert%E2%80%93van_Kampen_theorem

I am aware of "natural transformations" in category theory. However, I have trouble figuring out in this context why is $k$ called a "natural morphism"?

I can't figure out which are the "functors" and the "objects" in this case, to qualify $k$ as a natural transformation.

Thanks a lot.

yoyostein
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    By "natural" they mean canonical, which can be interpreted as being "independent of choices." This means if you were to pick a different $X, U_1, U_2$ with the given properties, the map $k$ would be defined in the same way. – Aurel Jul 10 '18 at 16:28
  • @Aurel I see, so it has nothing to do with natural transformations? – yoyostein Jul 10 '18 at 16:33
  • I don't know anything about homotopy. Isn't there any posibility for them to be functors? You must check if they are functors but imagine that we have $F$ a functor that to each pair of topological subspaces of $X$ it assigns the pushout of their fundamental groups, while $G$ is the functor that to each pair of subspaces of $X$ it assigns the homotopy group of their union. – Dog_69 Jul 10 '18 at 16:39
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    "Naturality" in natural transformations also (loosely) means "independent of choice." The naturality square says precisely that the transformation behaves "the same way" throughout the whole category (i.e. "independent of the choice of objects & morphisms"). The terms "natural" and "canonical" are often used in this way. – Aurel Jul 10 '18 at 16:50
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    For a more precise discussion, see chapter 2 of Peter May's $\textit{A concise course in algebraic topology},$ which can be found here: https://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf – Aurel Jul 10 '18 at 16:59
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    It is precisely the "natural map" induced by a colimit diagram as in category theory. – Randall Jul 10 '18 at 17:30
  • It is also natural in that it is a natural transformation from a certain functor to another (say you go from the category of quadruplets $(X,U_1,U_2,x_0)$ satisfying the hypotheses of the theorem to the category of groups, then the associated $k$ will be natural I think) – Maxime Ramzi Jul 10 '18 at 20:29

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$k$ is a natural transformation, although it's annoying to spell out in full detail how. The relevant functors have domain the category of tuples consisting of a topological space $X$, two open subspaces $U_1, U_2$ of $X$ with nonempty intersection, and a basepoint $x_0 \in U_1 \cap U_2$ (I'll leave the description of the morphisms in this category as an exercise). The codomain is groups. The first functor is $\pi_1(U_1) \ast_{\pi_1(U_1 \cap U_2)} \pi_1(U_2)$. And the second functor is $\pi_1(X)$ ($x_0$ is the basepoint throughout).

But as in the comments, "natural" here is also being used in a somewhat more informal sense to refer to a map that exists for "universal reasons" in some sense, by which I mean because of universal properties. Namely:

The spaces $U_1, U_1 \cap U_2, U_2, X$ fit into a commutative square (in fact a pullback square), and by the universal property of pushouts, this induces a map $U_1 \sqcup_{(U_1 \cap U_2)} U_2 \to X$ which preserves basepoints. Applying $\pi_1$ to this map (and using the fact that functors $F$ always have the property that there is a natural map from $\text{colim} F(-)$ to $F(\text{colim}(-))$, again by the universal property of colimits) produces the desired map $k$.

A simpler example of a natural map in this sense, although it is also a natural transformation, is the fact that $U_1 \cap U_2$ admits two natural inclusions, one into $U_1$ and one into $U_2$, by the universal property of intersections.

It's actually fairly unfortunate that "natural" gets overloaded in this way because this second meaning is subtle and takes some getting used to, but this is standard usage in category theory. Maybe we should switch to "universal" instead or something.

Qiaochu Yuan
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