# Homotopy equivalence of total space of fibrations

I am reading Selick’s book “Introduction to homotopy theory”, and he says in Cor 7.1.5 that for a fibration $p: E \rightarrow B$ with fiber $F$ and $B$ is CW complex, it is equivalent that there exists homotopy retraction $r: E \rightarrow F$ s.t. $rj \simeq 1_F$ where $j:F \rightarrow E$ is inclusion map, and there exists $r: E \rightarrow F$ s.t. $(r, p): E \rightarrow F \times B$ is homotopy equivalence.

Just before stating this corollary he states that for fibrations $p:X\rightarrow B$ and $q : Y \rightarrow B$ with $B$ : CW-complex and for a map $f: X \rightarrow Y$, $f$ is homotopy equivalence if and only if $f$ restricts homotopy equivalence on each fiber of $p$ and $q$.

I understand that $(r, p)$ is homotopy equivalence, then there exists homotopy retraction $r'$. But I have no idea why this converse holds. Does anybody prove that?

Thank you.

The corollary can only be true if we require the base space $B$ path connected, otherwise there may exist fibers of different homotopy type.

Let $F = p^{-1}(b)$ be the fiber over some $b$. We claim that the following are equivalent:

$(1)$ There exists a homotopy retraction $r : E \to F$.

$(2)$ There exists a map $r : E \to F$ such that $(r,p) : E \to F \times B$ is a fiber homotopy equivalence.

For each map $r : E \to F$ and each $x \in B$ define $r_x = r \mid_{F_x}: F_x = p^{-1}(x) \to F$. What we know is

(a) $r$ is a homotopy retraction if and only if $r_b = r \circ j \simeq 1_F$.

(b) $(r,p)$ is a fiber homotopy equivalence if and only if all $r_x$ are homotopy equivalences.

$(2) \Rightarrow (1)$: By (b) $r_b$ is a homotopy equivalence. Let $\sigma$ be a homotopy inverse for $r_b$. Define $r' = \sigma \circ r$. Then $r'_b = \sigma \circ r_b \simeq 1_F$. i.e. $r'$ is a homotopy retraction.

$(1) \Rightarrow (2)$: By (b) it suffices to show that all $r_x$ are homotopy equivalences. By (a) we know that $r_b$ is one. Let $u : I \to B$ be a path such that $u(0) = b, u(1) = x$. Form the pullback of $(u,p)$. This yields a fibration $p' : E' \to I$ and a map $u' : E' \to E$ such that $u' \circ p = p' \circ u$. The fibers $F'_t = (p')^{-1}(t)$ are mapped by $u'$ homeomorphically onto $F_{u(t)}$. Since $I$ is contractible, there exists a fiber homotopy equivalence $f : F'_0 \times I \to E'$. Define $i_t : F'_0 \to F'_0 \times I, i_t(y) = (y,t)$ and $i'_t : F'_0 \to F'_0 \times \{ t \}, i'_t(y) = (y,t)$. It suffices to show that $F'_0 \stackrel{i'_1}{\rightarrow} F'_0 \times \{ 1 \} \stackrel{f_1}{\rightarrow} F'_1 \stackrel{u'_1}{\rightarrow} F_x \stackrel{r_x}{\rightarrow} F$ is a homotopy equivalence. We have $r_x \circ u'_1 \circ f_1 \circ i'_1 = r \circ u' \circ f \circ i_1 \simeq r \circ u' \circ f \circ i_0 = r_b \circ u'_0 \circ f_0 \circ i'_0$, the latter being a homotopy equivalence.

• thank you for your polite answer. Can we give an example that (1) → (2) does not hold when the base space B is not path-connected?
– Sato
Jul 29 '18 at 8:39
• Take $B = \{ 0, 1 \}$, $E= \{ 0 \} \times \{ 0, 1 \} \cup \{ 1 \} \times \{ 0 \}$ and $p : E \to B, p((a,b)) = a$. Let $F = \{ 0 \} \times \{ 0, 1 \}$ be the fiber over $0$. Then $r : E \to F,r((a,b)) = b$ is a retraction, but $E$ and $B \times F$ are not homotopy equivalent. Jul 29 '18 at 8:47
• Now I understand completely. Thank you!
– Sato
Jul 29 '18 at 8:54