Ytukyg

This is a theorem from Tammo tom Dieck's Algebraic Topology:

While it has a direct proof, the author gives a more formal proof in the problems:

By pullback I suppose he means a diagram
$require{AMScd}$
begin{CD}
Ztimes_BE @>k>> E\
@V V q V @V V p V\
Z @>>f> B
end{CD}

However, I don't see how the existence of a lifting $Phi:Zto E$ is equivalent to the existence of a section $s:Zto Ztimes_BE$. (Even if you have a section $s$ and define $Phi:=ks$, it need not satisfy $Phi(z)=x$.) Also, he uses $E_b$ to denote that fibre, which suggests that it should be a subset of $E$. But in this pullback $q$ the fibre lives in $Ztimes_B E$.

I think I may have misunderstood the problem. Could you please tell me what he means by this? Thanks in advance!

Form the pullback $f^*E=Ztimes_BE$. This space sits in a strictly commutative diagram
$require{AMScd}$
begin{CD}
f^*E=Ztimes_BE@>k>> E\
@VVq V @VV p V\
Z @>f>> B
end{CD}
and has the universal property that for any space $A$ the projections $q,k$ induce a 1-1 correspondence

$${text{maps}: Arightarrow Ztimes_BE}leftrightarrows{text{pairs};(a:Arightarrow Z, b:Arightarrow E)mid fcirc a=pcirc b}.$$

Explicitly we send $F:Arightarrow Ztimes_BE$ to the pair $(qcirc F,kcirc F)$. The pullback is characterised by this universal property and so is unique up to homeomorphism. If you like you can take the standard model

$$Ztimes_BE={(z,e)in Ztimes Emid f(z)=p(e)}subseteq Ztimes E,$$

with the maps $q,k$ induced by projection onto each factor. In this way we make explicit the inverse to previous function: if $a:Arightarrow Z$, $b:Arightarrow E$ satisfy $fcirc a=pcirc b$ then we define $atimes_Bb:Arightarrow Ztimes_BE$ by $(atimes_Bb)(x)=(a(x),b(x))$, $xin A$. One check that this is well-defined and continuous.

In this way we get the correspondence between liftings of $f$ and sections of $q:Ztimes_BErightarrow Z$. If $Phi:Zrightarrow E$ lifts $f$ through $p$ then it satisfies $pcirc Phi=f$, and so the pair $(id_Z,Phi)$, which lives on the right-hand side of the previous correspondence, induces a map $s_Phi:Zrightarrow Ztimes_BE$ which will satisfy

$$qcirc s_Phi=id_Z,qquad kcirc s_Phi=Phi.$$

In particular the first of these conditions shows that $s_Phi$ is a section of $q$.

On the other hand, if $s:Zrightarrow Ztimes_BE$ is a section of $q$ then $Phi_s=kcirc s:Zrightarrow E$ lifts $f$. To wit,

$$pcirc Phi_s=pcirc (kcirc s)=(pcirc k)circ s=(fcirc q)circ s=fcirc (qcirc s)=fcirc id_z=f.$$

As for the condition $Phi_s(z)=x$ we observe that the fibre of $q$ over $a$ is homeomorphic to the fibre of $p$ of $f(z)$, the maps in the defining commutative square inducing the isomorphism. In fact using the explicit model gives us

$$(Ztimes_BE)_z={(z,e)mid f(z)=p(e)}cong {z}times E_{f(z)}.$$

Thus the requirement $Phi_s(z)=x$ is equivalent to the condition $s(z)=(z,x)$. From here one proceeds directly as in the given proof of (3.5.2) to get the result.

However, I think you are after a more extended discussion, especially relating to the exercise [3.5.1]. For this, I belive that tom Dieck is interested in the case that both $B$ and $Z$ are path connected, locally path connected and semi-locally simply connected, so I will assume these additional hypotheses.

Thus by (3.3.2) Classification I, and (3.4.1), the category of coverings $Cov_Z$ of $Z$ is equivalent to the category of $pi_zmbox-Set$ of (left) $pi_z=pi_1(Z,z)$-sets, where $zin Z$ is a chosen element. Above we have produced a pullback covering $q:f^*E=Ztimes_BErightarrow Z$ over $Z$, and this therefore corresponds to some $pi_z$-set which we'll denote $f^*mathcal{E}$. In fact we can see explicitly what this set is by what I have written above. The covering $p:Erightarrow B$ corresponds in $pi_{f(z)}$-set to the $pi_1(B,f(z))$-set given by the fibre $E_{f(z)}$. Therefore its pullback $f^*mathcal{E}$ should correspond to the $pi_z$-set given by the fibre

$$(f^*E)_zcong {z}times E_{f(z)},$$

and $pi_1(Z,z)$ should act on this set by the pullback action, that is,

$$alphacdot (z,e)=(z,(f_*alpha)cdot e),qquad (z,e)in f^*E,alphainpi_1(Z,z).$$

Now we can view a section $s:Zrightarrow f^*E$ of the pullback as a morphism of covering spaces
$require{AMScd}$
begin{CD}
Z@>s>> f^*E\
@VV= V @VV q V\
Z @>=>> Z,
end{CD}
and this corresponds to a morphism $mathfrak{s}:{z}rightarrow f^*mathcal{E}$ in $pi_z$-Set . Here I have used that the trivial covering $id_Z:Zrightarrow Z$ with one-point fibres corresponds to the trivial one-point $pi_z$-set ${z}$. Such a $pi_1$-equivariant morphism $mathfrak{s}$ must take the $pi_z$-fixed element $z$ into an element of $f^*mathcal{E}$ which is fixed under the pullback $pi_z$-action. This is exactly the statement that $E_{f(z)}$ has a fixed point under the $pi_1(Z,z)$-action.

On the other hand, if $kin f^*mathcal{E}$ is fixed by the pullback $pi_z$-action, then $mathfrak{s}:{z}rightarrow f^*mathcal{E}$, $ymapsto k$, is well-defined in $pi_z$-Set and so gives rise to a section $s:Zrightarrow f^*E$ in $Cov_Z$ since these categores are equivalent.

Finally, according to (3.2.8) the isotropy subgroup of $xin E_{f(z)}$ is exactly the image of $p_*:pi_1(E,x)rightarrow pi_1(B,f(z))$, so putting everything together, $(f^*E)_zcong{z}times E_{f(z)}$ will have a fixed point under the $pi_1(Z,z)$ action whenever $f_*(pi_1(Z,z))subseteq p_*pi_1(E,x)$. Hence the conditions are sufficente to choose a section $zmapsto (z,e)$ of $f^*E$, which by the previous is equivalent to a lifting $Phi:Zrightarrow E$, satisfying $Phi(z)=x$.

I just found a simpler proof that works without the hypothesis that $B$ is semi-locally simply connected (and therefore without using sophisticated results on the classification of covering spaces for such spaces). Moreover, I think it is closer to what the author had in mind.

First, for any covering $p:Eto B$, we have the transport functor $T_p:Pi(B)tomathsf{SET}$, where $T_p(b)=p^{-1}(b)$ for $bin B$, and for $v:Ito B$ the map $T_p(v):p^{-1}(v(0))to p^{-1}(v(1))$ is defined thus: given $ein p^{-1}(v(0))$ lift $v$ to $V:Ito E$ (i.e., $pV=v$) with initial condition $V(0)=e$ and set $[T_p(v)](e):=V(1)$. (This is introduced on page 67 of tom Dieck's book.)

The proof begins. We use the pushout
begin{CD}
Y @>k>> E\
@V V q V @V V p V\
Z @>>f> B
end{CD}
where $Y:={,(z,x)in Ztimes E:f(z)=p(x),}$ and $q,k$ are projections onto the coordinates. Then $q$ is also a covering. This induces transport functors $T_p:Pi(B)tomathsf{SET}$, $T_q:Pi(Z)tomathsf{SET}.$ Now observe that $T_q(z)={z}times p^{-1}(f(z))cong T_p(f(z))$ and this is a natural equivalence $eta:T_qsimeq T_pcircPi(f)$. (This is actually Problem 3.3.2 in disguise.) In particular we have a commutative diagram
begin{CD}
q^{-1}(z) @>eta_z>> p^{-1}(f(z))\
@V V T_q(v) V @V V T_p(fv) V\
q^{-1}(z) @>>eta_z> p^{-1}(f(z))
end{CD}
where $vinpi_1(Z,z)$. Thus $pi_1(Z,z)$ acts on $p^{-1}(f(z))$ if we identify $q^{-1}(z)$ and $p^{-1}(f(z))$ via $eta$.

The condition $f_*(pi_1(Z,z))subset p_*(pi_1(E,x))$ says $pi_1(Z,z)$ fixes $x$. So $q_*:pi_1(Y,(z,x))topi_1(Z,z)$ is an isomorphism. (tom Dieck, page 69, Proposition 3.2.8) If we take the component $Y_0$ of $Y$ containing $(z,x)$, then $q_0:=qmid_{Y_0}:Y_0to Z$ is a covering (tom Dieck, page 64, Remark 3.1.6; local path connectedness is used here) with fiber consisting of a single point, i.e., $q_0$ is a homeomorphism. (This gives the section mentioned in the exercise.) Finally, $kq_0^{-1}$ gives the desired lifting.