# the proof of lifting criterion

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?)

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.

• 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 $t$. – Paul Frost Oct 7 '19 at 11:46
• @PaulFrost I don't see how $h_t(0) = h_t(1) = x_0$ implies that $\tilde h_0$ is a loop at $\tilde x_0$. – feynhat Mar 18 at 11:16
• @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. – Paul Frost Mar 18 at 12:21
• @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). – feynhat Mar 18 at 12:52
• @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. – Paul Frost Mar 18 at 14:38

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.