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Since the functor $\pi_1$ preserves products, how can $\pi_1(T,t_0)\approx\pi_1(S^1,x_0)\times\pi_1(S^1,y_0)$ be not natural? Here $T$ is the torus.

Wikipedia gives an explanation, but I am still a bit confused.

It seems like that the product of several objects is natural but the converse is not. More precisely, if $Z$ is isomorphic to the product of $X,Y$ in a category $\mathcal{C}$ and $F$ is a functor from $\mathcal{C}$ to another category $\mathcal{D}$, which preserves products. Then the following two functors are naturally isomorphic: \begin{align*} \mathcal{C}\times\mathcal{C}\longrightarrow\mathcal{D} &\colon(X,Y)\mapsto F(X\times Y)\\ \mathcal{C}\times\mathcal{C}\longrightarrow\mathcal{D} &\colon(X,Y)\mapsto F(X)\times F(Y) \end{align*} However, there are no natural isomorphisms between $F$ and the following functor: \begin{equation*} \mathcal{C}\longrightarrow\mathcal{D}\colon Z\mapsto(X,Y)\mapsto F(X)\times F(Y) \end{equation*}

The culprit seems to be that the decomposition of $Z$ may not be unique (even when the components in different decompositions are isomorphic), thus the isomorphism \begin{equation*} Z\approx X\times Y \end{equation*} itself is usually not natural.

Am I right?

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Yes. But more importantly, $\pi_1 (T, t_0)$ and $\pi_1 (S^1, x_0) \times \pi_1 (S^1, y_0)$ are just groups, not functors, so it doesn't make sense to talk about natural transformations between them at all. – Zhen Lin Mar 5 '14 at 11:40
To me it is not clear what exactly you mean by saying that the isomorphism between $XZ$ and $X\times Y$ is not natural. You should check up on the precise meaning of natural in this context. – Tim Porter Mar 8 '14 at 7:14
up vote 5 down vote accepted

I don't agree with the explanation on Wikipedia. They write down an isomorphism, which is an instance of a natural isomorphism, and then claim that it is not natural. This is not correct. But their reasoning is that they - secretly - change the domain categories. More specifically: The natural isomorphism

$\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$

is natural in $X$ and $Y$, i.e. gives an isomorphism of functors

$\pi_1(- \times - ) \cong \pi_1(-) \times \pi_1(-) : \mathsf{Top}_*^2 \to \mathsf{Grp}.$

If $X,Y$ are two fixed spaces, we can consider the full subcategory $\{(X,Y)\}$ of $\mathsf{Top}_*^2$ and the isomorphism

$\pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y)$

is still natural, in this case with respect to all pairs of endomorphisms of $X$ and endomorphisms of $Y$.

Of course, there are endomorphisms of $X \times Y \in \mathsf{Top}_*$ which are not induced by endomorphisms of $(X,Y)$ in $\mathsf{Top}_*^2$. Wikipedia mentions $X=Y=S^1$ and a certain Dehn twist on $S^1 \times S^1$. But this is a completely different matter and has nothing to do with naturality! What they do is to remind the reader of the correct definition of naturality, but they don't give a proper counterexample, since they consider two functors on different domain categories. They compare $\pi_1(-) : \mathsf{Top}_* \to \mathsf{Grp}$ with $\pi_1(-) \times \pi_1(-) : \mathsf{Top}_*^2 \to \mathsf{Grp}$, which isn't possible.

For real examples of unnatural isomorphisms, see MO/139388.

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As Martin writes, one has to be clear about the categories one is dealing with

There is a functor $\pi_1: Top \to Gpd$, giving the fundamental groupoid of a topological space. This functor preserves products. (6.4.4 of Topology and Groupoids). A consequence (6.5.10) is a results on the effect on the morphisms of fundamental groupoids of homotopies of maps $X \to Y$.

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