Connectivity of fibers under fibration replacement

Question

Assume all the spaces mentioned below are simply connected CW complexes. Let $ f: X \to Y $ be a continuous surjctive map between CW complexes, where $ f $ is not necessarily a fibration. Assume that for every $ y \in Y $, the fiber $ f^{-1}(y) $ is $ k $-connected, i.e., $$ \pi_i(f^{-1}(y)) = 0 \quad \text{for all } i \leq k. $$

Consider a fibration $ p: E \to Y $ along with a weak equivalence $ i: E \to X $ such that the following diagram commutes (any such f can be replaced by such pair due to model category structure):

\begin{array}{ccc} X & \xrightarrow{i} & E \\ \downarrow{f} & & \downarrow{p} \\ Y & = & Y \end{array}

Does it follow that the fibers $ p^{-1}(y) $ for each $ y \in Y $ are also $ k $-connected, i.e., $$ \pi_i(p^{-1}(y)) = 0 \quad \text{for all } i \leq k \text{ and } y \in Y? $$

Answer 1

No, this isn't true.

Fix $k \geq 1$ and let $Y$ be the join $S^k \ast [0,1]$, which is homeomorphic to $D^{k+2}$. It is a simply-connected CW-complex and has trivial homotopy groups.

Let $X$ be the join $S^k \ast [0,1]^\delta$, where $[0,1]^\delta$ is the interval with the discrete topology. Then $X$ is a union of uncountably many $(k+1)$-cells - one per point of $[0,1]$ - along their commmon boundary $S^k$. As a consequence, $X$ is a simply-connected CW-complex with nontrivial $\pi_{k+1}$.

Let $f: X \to Y$ be the continuous bijection induced by the continuous bijection $[0,1]^\delta \to [0,1]$. The fibers of this map are all points and hence $(k+1)$-connected. However, for any weakly equivalent fibration $E \to Y$, the fibers have nontrivial $\pi_{k+1}$ by the long exact sequence $$ \dots \to \pi_{k+1}(Y) \to \pi_k(F) \to \pi_k(E) \cong \pi_k(X) \to \pi_k(Y) \cong 0 \to \dots $$ As a result, the fiber of the replacement is not $(k+1)$-connected.

Answer 2

Model categories don’t give you a map like $i$, they give you a map in the other direction.

If you do have a map like $i$ then the answer is yes because if $j: X \to E’$ is the equivalence to a fibration that model categories do give you, then $j \circ i $ is a fiberwise homotopy equivalence, so the fibers of $f \circ i $ are retracts (up to homotopy) of the fibers of $f$. Since $k$-truncated spaces are closed under retracts, the result follows.