Short exact sequence of topological groups which is split, but not topologically split

Question

Consider an exact sequence of locally compact groups $$1 \to A \overset{\iota}{\to} B \overset{\pi}{\to} C \to 1.$$ Naturally, I assume the homomorphisms are continuous. I should probably also assume that $\pi$ is a quotient map, in the topological sense, but let's leave this assumption out for now. Now, suppose that there exists a homomorphism $$r : B \to A$$ such that $r \circ \iota = \mathrm{id}_A$. Then, $b \mapsto (r(b), \pi(b))$ is a continuous, bijective homomorphism $B \to A \times C$.

In reasonable situations, one expects the inverse map $A \times C \to B$ to be continuous so that $B \cong A \times C$ as topological groups. Let us say that the sequence is topologically split if this holds. If the sequence, is topologically split, then there is a continuous homomorphism $s :C \to B$ such that $\pi \circ s = \mathrm{id}_C$. So, clearly a necessary condition for the sequence to be topologically split is

$\mathbf{(*)}$ There exists an open neighbourhood $U$ of $1$ in $C$ and a continuous function $s: U \to B$ (not assumed compatible with multiplicaiton!) with $s(1) = 1$ such that $\pi \circ s = \mathrm{id}_U$.

On the other hand, if $(*)$ holds, then $B$ is topologically split. Indeed, it suffices to check continuity of the inverse mapping $A \times C \to B$ on the neighbourhood $A \times U$ of $(1,1) \in A \times C$ and, in that neighbourhood, the inverse is given by the formula $(a,c) \mapsto \iota(a) s(c) \iota(r(s(c)))^{-1}$, whence is continuous. So, we have

Proposition: Consider an exact sequence of locally compact groups $1 \to A \overset{\iota}{\to} B \overset{\pi}{\to} C \to 1$ and suppose there is a (continuous) homomorphism $r : B \to A$ with $r \circ \iota = \mathrm{id}_A$. Then, the sequence is topologically split if and only if $(*)$ holds i.e. if and only if $\pi$ admits a continuous section, defined on a neighbourhood of $1 \in C$.

My question is as follows:

Question: Can someone think of an example where there is a continuous, retracting morphism $r : B \to A$, but the sequence is not topologically split?

Equivalently, by the above proposition, we need to take an example where $B$ is a not a locally trivial $A$-bundle over $C$. There are easy examples of this phenomenon: Take $A = \{\pm 1\}^\mathbb{N}$, the infinite product of 2-element groups (a Cantor set); $B = C =\mathbb{T}^\mathbb{N}$, the infinite product of the circle group; $\iota$ the evident inclusionl and $\pi$ the entrywise squaring map. This gives a short exact sequence $$1 \to \{\pm 1\}^\mathbb{N} \overset{\iota}{\to} \mathbb{T}^\mathbb{N} \overset{(z_i) \mapsto (z_i^2)}{\to} \mathbb{T}^\mathbb{N} \to 1$$ where $B$ is not an $A$-bundle over $C$, because $A$ is not locally-connected, and $B$ is. This example does not help with the question, because there is not retraction $B \to A$, but it does show that the condition $(*)$ is not automatic. The point is to show $(*)$ is still not automatic, even if there is a retraction $B \to A$.

Answer 1

I though about this some more, and I think that the answer is, if there is a continuous retracting homomorphism $B \to A$, then the sequence is topologically split. There's nothing particularly noteworthy about the argument, one just has to fiddle around with point-set topology. Also, the claim seems to go through for arbitrary topological groups, so the assumption of locally compact groups is not relevant.

First let's recall some general nonsense about topological groups.

Proposition: Let $G$ be a topological group, $N$ a normal subgroup and $\pi : G \to G/N$ the canonical projection. Give $G/N$ the quotient topology. Then,

1. $\pi : G \to G/N$ is actually an open map.
2. $G/N$ is a topological group.

Proof:

1. Suppose $U \subseteq G$ is open. Note that $\pi^{-1}(\pi(U)) = \bigcup_{x \in N} xU$ is open, so $\pi(U)$ is open by definition of the quotient topology.
2. Since a product of two open maps is open, $\pi \times \pi : G \times G \to G/N \times G/N$ is open and, in particular, a quotient map. So, by a property of quotient topologies, a map $f:G/N \times G/N \to Y$ is continuous if and only if $f \circ (\pi \times \pi) : G \times G \to Y$ is continuous. In particular, taking $f$ to be the multiplication map $G/N \times G/N \to G/N$, we need that $(x,y) \mapsto \pi(x)\pi(y) = \pi(xy)$ is continuous, which it is because it is composition of the multiplication map $G \times G\to G$ with $\pi$.

With these basic results recalled, we come to the main claim.

Propositon Let $G$ be a topological group, $N$ a normal subgroup, and $\pi : G \to G/N$ the standard projection. Suppose there exists a continuous homomorphism $r : G \to N$ such that $r(x)=x$ for all $x \in N$. Then, the continuous, bijective homomorphism $(r,\pi) : G \to N \times (G/N)$ has a continuous inverse, i.e $G \cong N \times (G/N)$ as topological groups.

Proof: Let $s = [\pi|_{\ker(r)}]^{-1}$ so that the inverse map is $(x,y) \mapsto x s(y) : N \times (G/N) \to G$. Clearly, to get the continuity, we just need to show that $s$ is continuous or, equivalently, that $\pi|_{\ker(r)} : \ker(r) \to G/N$ is an open map. To this end, take an arbitrary open set in $\ker(r)$, say $U \cap \ker(r)$ where $U$ is open in $G$. By the above proposition, $\pi(U)$ is open, but we have the problem that, quite possibly, $\pi(U \cap \ker(r)) \subsetneq \pi(U)$. We define a replacement for $U$ which does not have this problem: $$U' = \{ x \in G: x r(x)^{-1} \in U\}.$$ Clearly $U'$ is open in $G$ and, because $xr(x)^{-1} = x$ when $x \in \ker(r)$, we have $U' \cap \ker(r) = U \cap \ker(r)$. Also, for any $x \in U'$, we have $\pi(x) = \pi(x r(x)^{-1})$ (since $r(x)^{-1} \in N = \ker(\pi)$) where $x r(x)^{-1} \in U \cap \ker(r) = U' \cap \ker(r)$ (since $r(x r(x)^{-1}) = r(x)r(x)^{-1} = 1$), so $\pi(U \cap \ker(r)) = \pi(U' \cap \ker(r)) = \pi(U')$ is open in $G/N$, as claimed.

So yeah, that's pretty much the whole story I guess. Note, in the original post, I wrote "I should probably also assume that $\pi$ is a quotient map, in the topological sense, but let's leave this assumption out for now". I wrote this because this comes automatically in the case where there is a local section $C \to B$, but in hindsight I think one should probably just ignore this daft comment. I definitely want the 3rd group to be the quotient of the 2nd, in every sense.