Equivalence of contractible and trivial fibrations

Question

Let $B$ be a connected space. I am wondering, which of the following are equivalent, and under which conditions?

1. $B$ is weakly contractible.
2. $B$ is contractible.
3. All fibrations $E \rightarrow B$ are homotopy equivalent to a principal fibraiton $E' \times B \rightarrow B$, by a homotopy which commutes with the fibrations.
4. All fibrations $E \rightarrow B$ are weakly equivalent to a principal fibration $E' \times B \rightarrow B$, by a homotopy which commutes with the fibrations.

I am particularly interested in (3) implies (2) and (4) implies (1).

Answer 1

I will show the implications $2\Leftrightarrow3$ and $4\Rightarrow 1$. It is clear that $2\Rightarrow 1$ nonreversibly, and that $3\Rightarrow 4$

Let $B$ be a space.

Claim: $B$ is contractible if and only if every Hurewicz fibration $E\rightarrow B$ is fibre-homotopically trivial.

Proof: If $B$ is contractible then it is standard that every fibration $p:E\rightarrow B$ is fibre-homotopy equivalent to a product $B\times F$, where $F=p^{-1}(b_0)$, with $b_0\in B$ some chosen point. A proof of this can be found in Spanier's book Algebraic Topology on pg. 102.

Conversely assume that each fibration over $B$ is fibre-homotopy equivalent to a product. Choose a point $b_0\in B$ and let $e:PB\rightarrow B$ be the path-space fibration based at $b_0$. Thus $PB$ is the space of maps $\ell:I\rightarrow B$ satisfying $\ell(0)=b_0$, and the map $e$ is defined by $e(\ell)=\ell(1)$. The fibre of $e$ over $b_0$ is the loop space $\Omega B$. By assumption $PB\simeq B\times\Omega B$ over $B$. This implies that $B$ is a retract of $PB$. But $PB$ is contractible, so therefore so is its retract $B$. $\square$

For an example of a weakly contractible space over which there are homotopically non-trivial fibrations we can take the Warsaw circle $\mathcal{W}$. This space is weakly contractible but not contractible. In fact $[\mathcal{W},S^1]\cong \mathbb{Z}$. Since $\mathcal{W}$ is not contractible, the path-space fibration cannot be fibre-homotopically trivial. For a similar example see here.

Now, I can show that $4\Rightarrow 1$. I discuss the reason why you should not expect a converse below.

Claim: If each fibration over $B$ is weakly-fibre-homtopically trivial, then $B$ is weakly contractible.

Proof: We can again take the path-space fibration to get that $0=\pi_kPB\cong\pi_k(B\times\Omega B)\cong \pi_kB\oplus\pi_{k+1}B$. i.e. that $\pi_kB=0$ for each $k\geq0$. $\square$

Note that I have not even needed a fibrewise assumption for the proof. What I have proved is that if for any fibration $p:E\rightarrow B$ there is a weak equivalence $E\simeq_wB\times p^{-1}(b_0)$, then $B$ is weakly contractible.

In fact this last statement is reversible. For if $B$ is weakly contractible and $p:E\rightarrow B$ is a fibration, then by studying the long-exact homotopy sequence we see that the inclusion of any fibre $j:F\hookrightarrow B$ is a homotopy equivalence. Thus $B\times F\rightarrow\ast\times F\cong F\hookrightarrow E$ is a weak equivalence and we find that $E$ has the correct weak homotopy type. The problem is finding a fibrewise map to induce this equivalence.

This is why you should not expect $1$ to imply $4$. Basically the much weaker equivalence I have established in the last paragraph is what you should expect when working with weak equivalence. My guess is that $1\not\Rightarrow 4$, although I have no counterexample.

Ending with a remark, it is of course worth mentioning that if $B$ has CW homotopy type then there is no distinction between contractibility and weak contractibility. If $B$ does not have CW homotopy type, then even restricting to those fibrations with CW-type fibres and total space does not help. For instance in the example above of the Warsaw circle, both $P\mathcal{W}$ and $\Omega\mathcal{W}$ are contractible and so have CW homotopy type.