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In J. P. May's A Concise Course in Algebraic Topology, Chapter 2, He states the following lemma is "immediate from the universal property of products":

Lemma. For based spaces $X$ and $Y$, $\pi_1(X\times Y)=\pi_1(X)\times\pi_1(Y)$.

I know the more or less "standard" proof for this, but I don't see how I can use the universal property to deduce this. Although $\pi_1$ is a functor, but to my knowledge a functor need not preserve products.

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    $\begingroup$ The point is to prove that $\pi_1$ does preserve products, so it can't be as easy as using a general property of functors. But here's a hint: hom functors preserve products on their second entry, and $\pi_1$ is closely related to a hom functor. Try to see if you can work from there and update your question with your progress so people know how to help you. $\endgroup$ Apr 7 at 4:57
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    $\begingroup$ Category theory is called "abstract nonsense" not because it is nonsense, but because it is abstract. For concrete cases you must ultimately use some concrete property that applies to the specific case, for instance the definition of $\pi_1$. $\endgroup$
    – Trebor
    Apr 7 at 5:53

4 Answers 4

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Let $\pi:X\times Y \to X, \pi':X\times Y \to Y$ be the usual projections. Define $p_{1}:\pi_{1}(X\times Y) \to \pi_{1}(X)$ by $p_{1}[g] = \pi \circ g$ and $p_{2}:\pi_{1}(X \times Y) \to \pi_{1}(Y)$ by $p_{2}[g] = \pi' \circ g$. In the text that you reference, these are already available to you. Let $G$ be a group, and $f_{1}:G \to \pi_{1}(X), f_{2}:G \to \pi_{1}(Y)$ be homomorphisms.

Suppose there exists a homomorphism $\varphi:G \to \pi_{1}(X \times Y)$ such that $p_{1} \circ \varphi = f_{1}$ and $p_{2} \circ \varphi = f_{2}$. Let $x \in G$. Then, $(p_{1} \circ \varphi)(x) = f_{1}(x)$ and $(p_{2} \circ \varphi)(x) = f_{2}(x)$ so that $\pi \circ \varphi(x) = f_{1}(x)$ and $\pi' \circ \varphi(x) = f_{2}(x)$. And so $\varphi(x) = (f_{1}(x), f_{2}(x))$. This shows that $\varphi$ is unique, and you can verify that it is a well-defined homomorphism that sends $x \in G$ to an element of $\pi_{1}(X \times Y)$.

Therefore, $\pi_{1}(X \times Y)$ satisfies the universal property of the group product of $\pi_{1}(X)$ and $\pi_{1}(Y)$ i.e $\pi_{1}(X \times Y) \cong \pi_{1}(X) \times \pi_{1}(Y)$.

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  • $\begingroup$ I do not understand this proof. 1. You have a typo (it should be $G$ instead $Z$). 2. $\varphi(x) = (f_{1}(x), f_{2}(x))$ does not make sense because $\varphi(x) \in \pi_1(X \times Y)$ and $(f_{1}(x), f_{2}(x)) \in \pi_1(x) \times \pi_1(Y)$. $\endgroup$ Apr 16 at 9:02
  • $\begingroup$ One more typo: It should be $p_{1}[g] = [\pi \circ g]$. And what is $\pi \circ \varphi(x)$? You cannot compose a map between topological spaces with a group homomorphism. Certainly you mean $\pi_* \circ \varphi(x)$, but $\pi_* = p_1$, so you do not need $\pi$. $\endgroup$ Apr 17 at 10:03
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The abstract nonsense perspective is as follows: The fundamental group $\pi_1\colon h\mathbf{Top}_{\ast}\rightarrow\mathbf{Grp}$ is a functor on the homotopy category of pointed spaces that is represented by the pointed space $(S^1,1)$. Here, the fact that the functor is $\mathbf{Grp}$- rather than $\mathbf{Set}$-valued corresponds, by the Yoneda lemma, to a cogroup structure on $(S^1,1)$ in $h\mathbf{Top}_{\ast}$, which is given by the standard pinching map $S^1\rightarrow S^1\vee S^1$. In any case, representable functors do preserve products for more or less tautological reasons.

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    $\begingroup$ $\pi_1$ is not Ab-valued. $\endgroup$ Apr 17 at 10:07
  • $\begingroup$ @KritikerderElche Thanks for catching that. $\endgroup$
    – Thorgott
    Apr 17 at 13:03
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May introduces the fundamental group $\pi_1(Z,z)$ of a pointed space $(Z,z)$ as the set of equivalence classes of loops $f : I \to Z$ based at $z$. The equivalence realation is "path homotopy" which keeps the endpoints fixed.

It is easy to see that one can alternatively define $\pi_1(Z,z)$ as the set of pointed homotopy classes of pointed maps $f : (S^1,1) \to (Z,z)$.

Whatever our preference is, the proof of the Lemma is based on the universal property of the product in the category of pairs of topological spaces $(A,A_0)$ (i.e. $A_0 \subset A$) and continuous maps $f : (A,A_0) \to (B,B_0)$ of pairs (i.e. continuous maps $f : A\to B$ such that $f(A_0) \subset B_0$). Let us explain this without using much language of category theory.

  1. Define $\operatorname{Maps}(A, B)$ to be the set of continuous maps $f : A \to B$ and $\operatorname{Maps}((A,A_0), (B,B_0))$ to be the set of continuous maps $f : (A,A_0) \to (B,B_0)$ of pairs. In particular, $\operatorname{Maps}((A,A_0), (B,B_0)) \subset \operatorname{Maps}(A, B)$. Clearly $\operatorname{Maps}((A,\emptyset), (B,\emptyset)) = \operatorname{Maps}(A, B)$. This suggests to identify each space $A$ with the pair $(A,\emptyset)$.

  2. Define $(A,A_0) \times (B, B_0) = (A \times B, A_0 \times B_0)$ and $(A,A_0) \times I = (A \times I, A_0 \times I)$.

  3. A homotopy in $\operatorname{Maps}((A,A_0), (B,B_0))$ is a map $H \in \operatorname{Maps}((A,A_0) \times I, (B,B_0))$. The homotopies in $\operatorname{Maps}((A,A_0), (B,B_0))$ induce an equivalence relation $\simeq$ on $\operatorname{Maps}((A,A_0), (B,B_0))$. The equivalence classes are called homotopy classes. Let $[(A,A_0), (B,B_0)] = \operatorname{Maps}((A,A_0), (B,B_0))/\simeq$ denote the set of homotopy classes.

  4. Let $p_X : (X,X_0) \times (Y,Y_0) \to (X,X_0), p_X(x,y) = x$, and $p_Y : (X,X_0) \times (Y,Y_0) \to (Y,Y_0), p_Y(x,y) = y$, denote the projections.

The universal property of the product says that $$\phi : \operatorname{Maps}((A,A_0), (X,X_0) \times (Y,Y_0)) \to \operatorname{Maps}((A,A_0), (X,X_0)) \times \operatorname{Maps}((A,A_0), (Y,Y_0)),$$ $$\phi(\gamma) = (p_X \circ \gamma, p_Y \circ \gamma) \tag{1}$$ is a bijection. This is well-known in case $A_0 = X_0 = Y_0 = \emptyset$. It is easy to see that this transfers to the case of general pairs (simply consider the appropriate subsets).

What May means is

Theorem. $\phi$ induces a bijection $$\bar \phi : [(A,A_0), (X,X_0) \times (Y,Y_0)] \to [(A,A_0), (X,X_0)] \times [(A,A_0), (Y,Y_0)]$$ given by $\bar \phi([\gamma]) = ([p_X \circ \gamma], [p_y \circ \gamma])$.

Proof. Let $(A,A_0)' = (A,A_0) \times I$. All we need is that $$\phi : \operatorname{Maps}((A,A_0)', (X,X_0) \times (Y,Y_0)) \to \operatorname{Maps}((A,A_0)', (X,X_0)) \times \operatorname{Maps}((A,A_0)', (Y,Y_0))$$ is a bijection. This implies that $\gamma \simeq \gamma'$ iff $p_X \circ \gamma \simeq p_X \circ \gamma'$ and $p_Y \circ \gamma \simeq p_Y \circ \gamma'$. Together with $(1)$ this proves the theorem.

The fundamental group of $(Z,z)$ is nothing else than $$\pi_1(Z,z) = [(I,\{0,1\}), (Z,\{z\})] . \tag{1}$$ Now the above theorem applies.

Of course we can also work with $$\pi_1(Z,z) = [(S^1,\{1\}), (Z,\{z\})] . \tag{2}$$

Update (due to Kritiker der Elche's comment).

The other answers used an abstract categorical approach. So let me explain what I did above in a more categorical framework.

In Chapter 2.6. May introdoced the concept of a product of a family of objects in a category $\mathscr C$ (as a special case of a limit). Let $X = (X_i)_{i \in I}$ be a family objects of $\mathscr C$. Rephrasing a little, a product of $X$ is an object $P$ together with morphisms $p_i : P \to X_i$ such that the function of sets $$\phi : \mathscr C(A, P) \to \prod_{i \in I} \mathscr C(A, X_i), \phi(f) = (p_i \circ f)_{i \in I}$$ is a bijection.

The categories $\mathbf{Top}$ of topological spaces and $\mathbf{Top^2}$ of pairs of topological spaces have products (Cartesian product) and so does the category $\mathbf{Top_*}$ of pointed topological spaces. The theorem above shows that this fact transfers to the homotopy categories $\mathbf{hTop}, \mathbf{hTop}^2$ and $\mathbf{hTop_*}$. The "product morphisms" in these categories are nothing else than the homotopy classes $[p_i] : \prod_{i \in} X_i \to X_i$ of the projections $p_i : \prod_{i \in} X_i \to X_i$. In other words, the homotopy functors $h : \mathbf{Top} \to \mathbf{hTop}$ etc. preserve products. This requires of course a (simple) proof; see above.

By $(1)$ and $(2)$ the fundamental group functor $\pi_1 : \mathbf{hTop_*} \to \mathbf{Grp}$ followed by the forgetful functor $U : \mathbf{Grp} \to \mathbf{Set}$ is a $\operatorname{hom}$-functor and thus preserves products by definition.

In other words, the group homomorphism $$\psi : \pi_1\left(\prod_{i \in I} (X_i,x_i)\right) \to \prod_{i \in I}\pi_1(X_i,x_i), \psi(c) = ((p_i)_*(c))_{i \in I}$$ is a bijection on the level of underyling sets and thus a group isomorphism

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    $\begingroup$ Actually you proved step 1 in Nico's answer (in a more general form for pairs). The forgetful functor $U : Grp \to Set$ is essentially irrelevant. One has $\bar \phi(c) =((p_X)_*(c), (p_Y)_*(c))$ which is known to be a group homomorphism, and you have shown that it is a bijection. $\endgroup$ Apr 18 at 8:25
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You need the following facts.

  1. Products in $Top_*$ are products in the homotopy category $hTop_*$.
  2. The functor $U\pi_1: hTop_* \to Set$ is representable.
  3. Representable functors preserve products.
  4. The functor $U: Grp \to Set$ is monadic and hence creates limits.

The last point means that you can check that a diagram is a product in $Grp$ by applying the forgetful functor and checking it in $Set$.

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    $\begingroup$ Practically the same as Thorgott's answer a week ago. $\endgroup$ Apr 17 at 12:28
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    $\begingroup$ @KritikerderElche idk, representable functors into Set preserve products for "more or less tautological reasons". the addition is that Grp -> Set reflects limits. $\endgroup$
    – Nico
    Apr 17 at 13:07

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