infinite product of $1$-sphere has no universal covering space

Question

This is example 16 in Chapter 2, Section 5 of Spanier's Algebraic Topology. It states that: Any infinite product of $1$-sphere has no universal covering space. A universal covering space, as far as I know, need not to be simply connected. One counter example is offered in the same section of the book as Example 18. Any help would be appreciated.

The definition in Spanier for a universal cover is: If $p \colon \tilde{X} \to X$ is a universal cover. Then for any covering projection $q \colon \tilde{X}^\prime \to X$, there is $f \colon \tilde{X} \to \tilde{X}^\prime$ such that $q \circ f = p$. I do agree that a simply connected covering space of $X$ is an universal covering space of $X$, but I don't think the converse hold.

Answer 1

The problem is that if $\mathcal{A}$ is infinite, then the infinite product $\prod_{\mathcal{A}} S^1$ is not semi-locally simply connected. Compare this to:

Spanier, Corollary 14, pg. 83: A connected, locally path-connected space $X$ has a simply connected covering space if and only if it is semi-locally simply connected.

Clearly $\prod_{\mathcal{A}} S^1$ is connected and locally path-connected. Thus if it is not semi-locally simply connected, then it cannot admit a simply connected covering space. On the other hand, I will argue below that any universal covering space of $\prod_{\mathcal{A}} S^1$ must necessarily be simply connected. The contradiction establishes that this space admits no universal covering.

$\prod_{\mathcal{A}} S^1$ is not semi-locally simply connected: The condition for a space $X$ to be semi-locally simply connected is for each point $x$ to have an open neighbourhood $U\subseteq X$ such that the homomorphism $\pi_1U\rightarrow \pi_1X$ induced by the inclusion is trivial. Now the infinite product $\prod_{\mathcal{A}} S^1$ has the coarsest topology such that all projections $pr_a:\prod_{a\in\mathcal{A}} S^1\rightarrow S^1$ are continuous. When $\mathcal{A}$ is finite this coincides with the box topology, namely the topology generated by the sets $\{\prod_{a\in\mathcal{A}}U_a\mid U_a\subseteq S^1_a\; \text{open}\}$. However when $\mathcal{A}$ is not finite, then the box topology is strictly finer than the product topology, and the two do not coincide.

Unpacking the definitions, the product topology is generated by arbitrary unions and finite intersections of the form $pr_i^{-1}(U_i)=\{(x_a)_{a\in\mathcal{A}}\in\prod_{a\in\mathcal{A}}S^1\mid x_i\in U_i\}$. Clearly any finite intersection $U$ of such subsets will have non-trivial first homotopy, and the inclusion $\pi_1U\rightarrow \pi_1(\prod_{a\in\mathcal{A}}S^1)$ will not be trivial. Hence $\prod_{a\in\mathcal{A}}S^1$ is not semi-locally simply connected.

$\prod_\mathcal{A} S^1$ has no non-simply connected universal cover: Suppose $p:X\rightarrow \prod_{a\in\mathcal{A}}S^1$ is a connected covering. Then $p_*:\pi_1X\rightarrow \pi_1(\prod_{a\in\mathcal{A}}S^1)\cong \prod_{a\in\mathcal{A}}\pi_1S^1\cong \prod_{a\in\mathcal{A}}\mathbb{Z}$ is monic. Note that the target group is abelian, so conjugacy is trivial. If $p_*(\pi_1X)=0$ then $X$ is simply connected and therefore cannot be universal by the above. Assume therefore that there a non-trivial element $\alpha\in p_*(\pi_1X)\subseteq \prod_{a\in\mathcal{A}}\pi_1S^1$. Then there exists a covering space $q:Y\rightarrow \prod_{a\in\mathcal{A}}S^1$ such that $\alpha\not\in q_*(\pi_1Y))$. These are easily constructed, see example 8, pg. 69. Now $\prod_{a\in\mathcal{A}}S^1$ is connected and locally path connected, so theorem 2, pg. 79 applies and states that there can be no morphism $X\rightarrow Y$ in the category of connected covering spaces. Hence $X$ cannot be univeral. Therefore the only covering of $\prod_{a\in\mathcal{A}}S^1$ that could be universal would be simply connected. But such an object does not exist.

Answer 2

Another argument using the free action of $\pi_1$ on the fibers of a universal cover can be given as follows:

Suppose $Y:=\prod_{i=1}^\infty S^1$ has a universal covering space $p:X\rightarrow Y$. Fix $x_0\in S^1$. Consider the element $\overline x=(x_0,\ldots,x_0,\ldots)$ of $Y$. There is an open nhood $U$ of $\overline x$ and some open $\tilde U$ of $X$, such that $p\upharpoonright\tilde U$ is a homeomorphism onto $U$. We may assume $U$ has the form $$\prod_{i=1}^{n-1}U_i\times\prod_{i=n}^\infty S^1,$$ where each $U_i$ is an open nhood of $x_0$. Now consider the loop $\alpha:S^1\rightarrow \prod_{i=0}^\infty S^1$ given by $x\mapsto (x_0,\ldots,x_0,x,x_0,\ldots)$, where the $``x"$ appears in the $n$th position. Then $\alpha$ is a loop of $U$. As $p\upharpoonright\tilde U$ is a homeomorphism we get that $\alpha$ lifts via $p$ to a loop $\gamma$ of $X$ contained in $\tilde U$.

However, $\alpha$ is a non-trivial loop of $Y$, and thus as $p:X\rightarrow Y$ is the universal cover of $Y$, we must have that the lift of $\alpha$ is not a loop.

This contradiction shows $Y$ cannot have a universal covering.