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If I have two (Serre-)fibrations over the same base, and a weak equivalence of the total spaces that is also a map over the base, could I hope that the induced map on the spaces of sections would also be a weak equivalence? If I restrict to CW-spaces, this is fairly easy, but I don't really know how to handle the space of sections if only weak equivalences are available.

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2 Answers 2

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This is not true in general, unless you assume the base is sufficiently nice (eg a CW-complex). Here is a counter-example.

Let $B = \mathbb{Q}$, the rationals with its topology as a subspace of the reals $\mathbb{R}$. Let $E_2 = \mathbb{Q}$ as well and let $E_1 = \mathbb{Q}_\delta$, the rationals with the discrete topology.

Then $E_2 \to B$ is the identity map (a homeomorphism) and is a Serre fibration. The set-theoretic identity map $E_1 \to B$ is also continuous. Since any map from a disk into $E_1$ or $B$ factors through a constant map, you can check that it is also a Serre fibration. The map $E_1 \to E_2$ is also a weak homotopy equivalence and a map over $B$, as you require.

However the space of sections of $E_2$ is the one point space, while the space of sections of $E_1$ is empty. In particular they are not weak homotopy equivalent.

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  • $\begingroup$ thanks for the example. Your also saying that for the base a CW-complex, we should be fine? would be much appreciated if you could point me towards a reference or a way to see this. I still find these things somewhat inaccessible $\endgroup$
    – Amueller
    Commented Mar 9, 2018 at 17:02
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As a complement to Chris Schommer-Pries answer above, the result holds when the base is a CW complex, as can be seen by a (possibly transfinite induction) on the cells. Here's a sketch.

Clearly, the result is true when $B$ is a point.

For a subcomplex $A\subset B$ let $\Gamma(E_i|A)$ be the space of sections of the restriction $E_{i|A} \to A$ for $i=1,2$. Suppose $A\subset C \subset B$, where $C = A\cup D^j$ is the result of attaching a single cell to $A$. Then we have a pullback square $\require{AMScd}$ $$ \begin{CD} \Gamma(E_i|C) @>>> \Gamma(E_i|A)\\ @VVV @VVV \\ \Gamma(E_i|D^j) @>>> \Gamma(E_i|S^{j-1}) \end{CD}$$ which is also a homotopy pullback because the horizontal arrows are Serre fibrations. This gives a long exact sequence in homotopy groups $$ ...\to \pi_n(\Gamma(E_i|C)) \to \pi_n(\Gamma(E_i|D^j)) \oplus \pi_n(\Gamma(E_i|A)) \to \pi_n(\Gamma(E_i|S^{j-1}))\to ... $$ (where I am being a bit sloppy with the notation for $n \le 1$). The square for $i=1$ maps to the square for $i=2$.

Inductively, assume the result is true for $A$. The result is clearly true for $S^{j-1}$ and $D^j$ since in these cases where are dealing with a homotopically trivial fibration of the form $F \times X \to X$ where $X= D^j$ or $S^{j=1}$. Then by the five lemma applied to the long exact sequence, we obtain the result for $C$. This gives the inductive step.

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    $\begingroup$ In the case when the base is a CW complex, maybe the result follows from model categorical nonsense: $\mathrm{map}_B(B, E)$ is already the `derived' mapping space since the source is cofibrant and the target is fibrant? but maybe I'm missing something $\endgroup$ Commented Mar 11, 2018 at 16:50
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    $\begingroup$ You are assuming that the person who asked the question knows mod cats. Of course you are correct. I decided to provide him with a direct sketch, since it’s more elementary. $\endgroup$
    – John Klein
    Commented Mar 11, 2018 at 17:33
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    $\begingroup$ I'm all for elementary :) just asking to make sure I wasn't unaware of some subtlety $\endgroup$ Commented Mar 11, 2018 at 17:46
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    $\begingroup$ much appreciated, thank you. the model categorical viewpoint also makes sense to me, it didnt occur to me to consider it that way... much to learn ; ) $\endgroup$
    – Amueller
    Commented Mar 14, 2018 at 12:33

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