25

The answer is no. If $L$, $L^\prime$ are 3-dimensional lens spaces and $S^1\times L$ is diffeomorphic to $S^1\times L^\prime$, then the covering space of $S^1\times L$ corresponding to the torsion subgroup defines an
h-cobordism between $L$ and $L^\prime$ (we have embeddings of L and L′ in the covering space with disjoint images, and the images bound an h-...

ag.algebraic-geometry dg.differential-geometry at.algebraic-topology differential-topology smooth-manifolds

16

Even if you're only interested in say cohomology with coefficients in the constant sheaf, working with constructible sheaves gives you extra flexibility and is more amenable to inductive proofs.
Here is a basic theorem in the topology of algebraic varieties one of whose proofs could serve as a motivation. I discussed it in some MO answer, maybe I'll link it ...

9

The answer to your question is known when $\mathcal{S}$ is a symmetric monoidal $\infty$-groupoid, by work of Galatius-Madsen-Tillmann-Weiss.
In other words: we understand invertible $2$-dimensional TFTs.
This might seem like a somewhat silly case, but it's actually a very useful way of testing conjectures.
Below I'll sketch how this implies that the answer ...

8

Think about the case where $\pi_0(G)=\mathbb{Z}$, so $B(\pi_0(G))=S^1$, so we have a fibre bundle $BG_0\to BG\to S^1$. In this case $G$ is always a semidirect product formed using an automorphism $\alpha$ of $G_0$. By considering the preimages of the the complements of two points in $S^1$, we can express $BG$ as $U\cup V$, where $U$ and $V$ are each open ...

5

There is a notion of the étale homotopy type of a (Grothendieck) topos, going back to Artin and Mazur (I think).
However, in classic "French" fashion they turned a theorem (in one setting) into a definition (in a more general setting): roughly speaking, two toposes have the same étale homotopy type if they have the same étale fundamental group and ...

5

For $n=1$, the answer to your question is negative, as explained by Gregory Arone in the comments.
In the cases $n\neq 1,2,4$, there is the following easy argument:
The long exact sequence of the fibration $S^{2n-1}\to BO(2n-1)\to BO(2n)$ induces an exact sequence
$$\pi_{2n}(BSO(2n))\xrightarrow{e} \mathbb{Z}\xrightarrow{[TS^{2n-1}]}\pi_{2n-1}(BSO(2n-1)),$$
...

5

Theorem 5.3 of
Daniel Fuentes-Keuthan, Modelling Connective Spectra via Multicategories, arXiv:1909.11148.
answers this question positively!

4

$\newcommand{\Z}{\mathbb Z}\newcommand{\cA}{\mathcal
A}\newcommand{\Sq}{\mathrm{Sq}}\newcommand{\CP}{\mathbb{CP}}\newcommand{\Ext}{\mathrm{Ext}}$It's possible to run
the Adams spectral sequence directly to show that the $2$-torsion subgroup of $\pi_7^s(\CP^2\wedge\CP^2)$ is isomorphic to $\Z/2$, so that
$\partial$ can't be an isomorphism. The calculation is ...

4

Another proof appears as Lemma 8.1.1.15 and Remark 8.1.1.16 of Igusa's Higher Franz-Reidemeister torsion. It uses spaces of graphs to verify that $|\Lambda|$ classifies oriented circle bundles.

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