# Why are the total spaces of two Serre fibrations equivalent when the bases and the fibers are equivalent?

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.

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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.
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 '14 at 0:18