Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $B$ is a pointed space and suppose $f\colon E\to B$ and $f\colon E'\to B$ are two Serre fibrations. Let moreover a map $g\colon E\to E'$ be given such that $f=f'\circ g$ which is a weak equivalence when restricted to the fibers $E_x$ and $E'_x$ of $E$ and $E'$ over the basepoint $x$ of $B$.

Eventually, I want $g$ to be a weak equivalence but perhaps I need $B$ to have trivial $\pi_0$.

By the long exact sequences from the Serre fibrations and the compatible maps inbetween, I got already that $\pi_k(g)$ is an isomorphism for $k\geq 2$ by the five-lemma. For $k=1$, there is a five-lemma for non-abelian groups which seems also to work if the very right entry $\pi_0(E_x)\cong\pi_0(E'_x)$ is only a pointed set.

My question concerns the end $$ \begin{array}{cccccccc} \cdots\to&\pi_1B &\to & \pi_0 E_x &\to & \pi_0E &\to & \pi_0 B\\ &\downarrow\cong && \downarrow\cong && \downarrow && \downarrow\cong\\ \cdots\to&\pi_1B &\to & \pi_0 E'_x &\to & \pi_0E' &\to & \pi_0 B\\ \end{array} $$ of the two horizontal long exact sequences. How to conclude that $\pi_0(g)$ is an isomorphism if $\pi_0B\cong *$? How to conclude that $\pi_0(g)$ is an isomorphism, if we have equivalence of the fibers for any basepoint $x$?

Edit: I am especially interested in the situation that all the spaces involved are CW complexes. Sorry, that I didn't mention this before.

share|improve this question

1 Answer 1

Assuming $B$ is connected, $\pi_0 E'_x\to \pi_0 E'$ is surjective, since all of $\pi_0E'$ is the inverse image of the distinguished point of $\pi_0 B$. Thus $\pi_0 g$ is surjective, by the usual diagram chase.

But it doesn't appear to me that $\pi_0 g$ has to be injective. As a counterexample, I propose to let $B$ be the interval $[0,1]$ with the cofinite topology and $E,E'$ be the projection from $B\times \{0,1\}$, $E$ with the product topology and $E'$ with the cofinite topology again. $E$ has the two path components $B\times \{0\}$ and $B\times \{1\}$ whereas $E'$ is connected, since every two open subsets have uncountable, in particular nonempty, intersection. On the other hand, the fibers of both $E$ and $E'$ are discrete two-point spaces.

Finally, define $g:E\to E'$ to be the identity on underlying sets, which is continuous because the cofinite topology is coarser than the product topology. This is a bijection, hence weak equivalence, on fibers, but maps both components of $E$ to the single component of $E'$. All that remains to check is that $E'$ is a fibration. It should suffice to lift homotopies of paths $I\to B$ to $E'$, which is simple: just apply the homotopy to the $B$ coordinate without changing the $\{0,1\}$ coordinate.

share|improve this answer
2  
Incidentally, I don't see how to put this argument through if $E,E',$ and $B$ are required to be Hausdorff. –  Kevin Carlson May 10 at 0:18
    
Kevin, thanks for the answer. Sorry not to have mentioned it, but I am interested in the case that all the spaces involved are CW complexes. –  jeffrey May 10 at 16:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.