An alternative, more formal proof of a path lifting criterion in tom Dieck's Algebraic Topology

Question

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} Z\times_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 $\Phi:Z\to E$ is equivalent to the existence of a section $s:Z\to Z\times_BE$. (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 $Z\times_B E$.

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

Answer 1

Form the pullback $f^*E=Z\times_BE$. This space sits in a strictly commutative diagram $\require{AMScd}$ \begin{CD} f^*E=Z\times_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

$$\{\text{maps}\: A\rightarrow Z\times_BE\}\leftrightarrows\{\text{pairs}\;(a:A\rightarrow Z, b:A\rightarrow E)\mid f\circ a=p\circ b\}.$$

Explicitly we send $F:A\rightarrow Z\times_BE$ to the pair $(q\circ F,k\circ F)$. The pullback is characterised by this universal property and so is unique up to homeomorphism. If you like you can take the standard model

$$Z\times_BE=\{(z,e)\in Z\times E\mid f(z)=p(e)\}\subseteq Z\times E,$$

with the maps $q,k$ induced by projection onto each factor. In this way we make explicit the inverse to previous function: if $a:A\rightarrow Z$, $b:A\rightarrow E$ satisfy $f\circ a=p\circ b$ then we define $a\times_Bb:A\rightarrow Z\times_BE$ by $(a\times_Bb)(x)=(a(x),b(x))$, $x\in A$. One check that this is well-defined and continuous.

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

$$q\circ s_\Phi=id_Z,\qquad k\circ s_\Phi=\Phi.$$

In particular the first of these conditions shows that $s_\Phi$ is a section of $q$.

On the other hand, if $s:Z\rightarrow Z\times_BE$ is a section of $q$ then $\Phi_s=k\circ s:Z\rightarrow E$ lifts $f$. To wit,

$$p\circ \Phi_s=p\circ (k\circ s)=(p\circ k)\circ s=(f\circ q)\circ s=f\circ (q\circ s)=f\circ id_z=f.$$

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

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

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 $\pi_z\mbox-Set$ of (left) $\pi_z=\pi_1(Z,z)$-sets, where $z\in Z$ is a chosen element. Above we have produced a pullback covering $q:f^*E=Z\times_BE\rightarrow Z$ 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 $p:E\rightarrow B$ 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

$$(f^*E)_z\cong \{z\}\times E_{f(z)},$$

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

$$\alpha\cdot (z,e)=(z,(f_*\alpha)\cdot e),\qquad (z,e)\in f^*E,\alpha\in\pi_1(Z,z).$$

Now we can view a section $s:Z\rightarrow f^*E$ 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 $id_Z:Z\rightarrow Z$ 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 $k\in f^*\mathcal{E}$ is fixed by the pullback $\pi_z$-action, then $\mathfrak{s}:\{z\}\rightarrow f^*\mathcal{E}$, $y\mapsto k$, is well-defined in $\pi_z$-Set and so gives rise to a section $s:Z\rightarrow f^*E$ in $Cov_Z$ since these categores are equivalent.

Finally, according to (3.2.8) the isotropy subgroup of $x\in E_{f(z)}$ is exactly the image of $p_*:\pi_1(E,x)\rightarrow \pi_1(B,f(z))$, so putting everything together, $(f^*E)_z\cong\{z\}\times E_{f(z)}$ 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 $z\mapsto (z,e)$ of $f^*E$, which by the previous is equivalent to a lifting $\Phi:Z\rightarrow E$, satisfying $\Phi(z)=x$.

Answer 2

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 $p:E\to B$, we have the transport functor $T_p:\Pi(B)\to\mathsf{SET}$, where $T_p(b)=p^{-1}(b)$ for $b\in B$, and for $v:I\to B$ the map $T_p(v):p^{-1}(v(0))\to p^{-1}(v(1))$ is defined thus: given $e\in p^{-1}(v(0))$ lift $v$ to $V:I\to E$ (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 $Y:=\{\,(z,x)\in Z\times E:f(z)=p(x)\,\}$ and $q,k$ are projections onto the coordinates. Then $q$ is also a covering. This induces transport functors $T_p:\Pi(B)\to\mathsf{SET}$, $T_q:\Pi(Z)\to\mathsf{SET}.$ Now observe that $T_q(z)=\{z\}\times p^{-1}(f(z))\cong T_p(f(z))$ and this is a natural equivalence $\eta:T_q\simeq T_p\circ\Pi(f)$. (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 $v\in\pi_1(Z,z)$. 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 $q_*:\pi_1(Y,(z,x))\to\pi_1(Z,z)$ is an isomorphism. (tom Dieck, page 69, Proposition 3.2.8) If we take the component $Y_0$ of $Y$ containing $(z,x)$, then $q_0:=q\mid_{Y_0}:Y_0\to Z$ 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.