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In Hatcher's book,the lifting criterion is stated as following:

Suppose given a covering space $p: (\tilde{X},\tilde{x}_0) \rightarrow (X,x_0)$ and a map $f: (Y,y_0) \rightarrow (X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\tilde{f}: (Y,y_0) \rightarrow (\tilde{X},\tilde{x}_0)$ of $f$ exists iff $f_*(\pi_1 (Y,y_0)) \subset p_*(\pi_1(\tilde{X},\tilde{x}_0))$.

The proof of "only if" part is obvious. I met with some problems when reading the proof the "if" part.

For any $y\in Y$,let $\gamma$ be a path in $Y$ from $y_0$ to $y$, then we can define $\tilde{f}(y)=\widetilde{f\gamma}(1)$.We need to show that the definition is well defined.Let $\gamma^{'}$ be another path in $Y$ from $y_0$ to $y$.Then $(f\gamma')\cdot(\overline{f\gamma})$ is a loop $h_0$ at $x_0$ with $[h_0]\in f_*(\pi_1 (Y,y_0)) \subset p_*(\pi_1(\tilde{X},\tilde{x_0}))$.*This means that there is a homotopy $h_t$ of $h_0$ to a loop $h_1$ that lifts to a loop $\tilde{h}_1$ in $\tilde{X}$ based at $x_0$.*How to deduce the above conclusion?

Applying the covering homotopy property to $h_t$ to get a lifting $\tilde{h}_t$.Since $\tilde{h}_1$ is a loop at $\tilde{x}_0$,so is $\tilde{h}_0$ .(Why?)

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Since $[h_0 ]\in p_{\ast }\pi_1 (\tilde{X} ,\tilde{x}_0 )$, it follows that there is a loop $\gamma_1\colon\mathbb{S}^1\to\tilde{X} $ such that $\gamma_1 (1)=\tilde{x}_0 $ and $p_{\ast} [\gamma_1 ]=[h_0 ]$. Consider the loop $h_1 =p\gamma_1 $, it represents the same homotopy class as $h_0 $, hence there is a homotopy $h_t $ between them.

For the second one, note that we have $h_t (1)$ and $h_t (0)$ constant loops, hence $\tilde{h}_t (1)$ and $\tilde{h}_t (0)$ satisfies $p\tilde{h}_t (i)\equiv h_t (i)$, $i=0,1$. Therefore $\tilde{h}_t (i)\in p^{-1} (h_t (i))$, which is discrete and therefore constant.

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  • $\begingroup$ You should write $\gamma_1 : I \to \tilde X$ because in Hatcher's book loops are defined as closed paths. And I think you shouldn't say $h_t(1)$ and $h_t(0)$ are constant loops (although you may regard $t \mapsto h_t(i)$ as a loop) because it is confusing. In fact we have $h_t(i) = x_0$ for all $t$. $\endgroup$ – Paul Frost Oct 7 '19 at 11:46
  • $\begingroup$ @PaulFrost I don't see how $h_t(0) = h_t(1) = x_0$ implies that $\tilde h_0$ is a loop at $\tilde x_0$. $\endgroup$ – feynhat Mar 18 at 11:16
  • $\begingroup$ @feynhat Although you should ask TheWildCat who wrote the answer, it is clear that the constant loop at $x_0$ lifts to a path in the fiber over $x_0$ which is discrete. Thus the lift is a constant loop. $\endgroup$ – Paul Frost Mar 18 at 12:21
  • $\begingroup$ @PaulFrost I know that the lift of constant loop is constant loop in the fiber, but $h_0$ is not a constant loop. See the construction of $h_0$ in the question, $h_0 = (f\circ\gamma')\cdot\overline{(f\circ\gamma)}$. (BTW, I did want to ask TheWildCat as well, but SE doesn't allow you to @ someone who hasn't participated in the comment section). $\endgroup$ – feynhat Mar 18 at 12:52
  • $\begingroup$ @feynhat Any comment to an answer is implicitly an @ to the person who gave the answer and occurs in his/her "Recent inbox messages". I did not claim that the TheWildCat's answer is perfect, I only commented that some points have to be clarified. $\endgroup$ – Paul Frost Mar 18 at 14:38
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Both of your questions can be answered by one simple fact that the elements of $p_*(\pi_1(\widetilde X, \widetilde x_0))$ precisely the classes of those loops in $(X, x_0)$ which lift to a loop in $(\widetilde X, \widetilde x_0)$. In fact this result is presented in Hatcher (the relevant part is in bold):

Proposition 1.31. The map $p_*: \pi_1(\widetilde X, \widetilde x_0) \to \pi_1(X, x_0)$ induced by a covering space $p : (\widetilde X, \widetilde x_0) \to (X, x_0)$ is injective. The image subgroup $p_*(\pi_1(\widetilde X, \widetilde x_0))$ in $\pi_1(X, x_0)$ consists of the homotopy classes of loops in $X$ based at $x_0$ whose lifts to $\widetilde X$ starting at $\widetilde x_0$ are loops.

The proof is given in Hatcher, but I present the proof of the part in bold for completeness.

Suppose $[\gamma] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$, then $[\gamma] = p_*[\widetilde \gamma_1]$, for some loop $\widetilde \gamma_1$ in $(\widetilde X, \widetilde x_0)$. So, we have $\gamma \simeq p \circ \widetilde \gamma_1 = \gamma_1$ (say). $\gamma$ and $\gamma_1$ are homotopic as paths, so their liftings will also be homotopic as paths (by homotopy lifting property), that is, $\widetilde \gamma_1 \simeq \widetilde \gamma$. Now, since $\widetilde \gamma_1$ is a loop, so, is $\widetilde \gamma$.

Conversely, suppose $\gamma$ is a loop in $(X, x_0)$ that lifts to a loop $\widetilde \gamma$ in $(\widetilde X, \widetilde x_0)$, this means that $\gamma = p \circ \widetilde \gamma$ or $[\gamma] = p_*[\widetilde\gamma]$. So, $[\gamma] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$.

Now, coming back to your questions, note that since $[h_0] \in p_*(\pi_1(\widetilde X, \widetilde x_0))$, $h_0$ lifts to loop.

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