1
$\begingroup$

The statement to prove given in Allen Hatcher's book Algebraic Topology is:

Given a map $F:Y\times I\rightarrow X$ and a map $\widetilde F:Y\times \{0\}\rightarrow \widetilde X$ lifting $\left.F\right|_{Y\times\{0\}}$, then there is a unique map $\widetilde F:Y\times I\rightarrow\widetilde X$ lifting $F$ and restricting to the given $\widetilde F$ on $Y\times\{0\}$.

Note that $I=[0,1]$, the unit interval. Also note that $(\widetilde X,p)$ is a covering space of $X$. My understanding of the first part of the proof of the existence is as follows (in my words):

Hatcher sets out to construct $\widetilde F:N\times I\rightarrow\widetilde X$ for some neighborhood $N$ of a given $y_0\in Y$. He first covers the compact space $\{y_0\}\times I$ by open products $N_t\times I_t$ containing $(y_0,t)$ and chooses finitely many of those already covering. Finite intersection of those $N_t$'s provides a single neighborhood $N$ of $y_0$, and $I$ can be partitioned into $0=t_0<...<t_m=1$ s.t. $F(N\times[t_i,t_{i+1}])$ is contained in an evenly covered neighborhood $U_i$.

So far I'm good. But then Hatcher makes an inductive construction (in my words):

Start with the given $\widetilde F$ on $N\times\{0\}$ and construct $\widetilde F$ on $N\times[0,t_i]$ inductively. Since $F(N\times[t_i,t_{i+1}])\subset U_i$ and the latter is evenly covered, there exists $\widetilde U_i$ containing $\widetilde F(y_0,t)$ mapping homeomorphically onto $U_i$ by $p$.

Then comes the line which I do not fully understand (quotation this time):

After replacing $N$ by a smaller neighborhood of $y_0$ we may assume that $\widetilde F(N\times\{t_i\})$ is contained in $\widetilde U_i$, namely, replace $N\times\{t_i\}$ by its intersection with $(\widetilde F|N\times\{t_i\})^{-1}(\widetilde U_i)$. Now we can define $\widetilde F$ on $N\times[t_i,t_{i+1}]$ to be the composition of $F$ with the homeomorphism $p^{-1}:U_i\rightarrow\widetilde U_i$. After a finite number of steps we eventually get a lift $\widetilde F:N\times I\rightarrow\widetilde X$ for some neighborhood $N$ of $y_0$.

QUESTION: Why do we need to replace $N$ with a smaller neighborhood here? Can anyone make that part become more clear to me?

$\endgroup$
  • $\begingroup$ You need $\widetilde{F}(N \times \{t_i\})$ to be contained in $\widetilde{U}_i$ so you can make use of the homeomorphism $p$. Otherwise it may contain elements in other components of the preimage of $U_i$. $\endgroup$ – Clive Newstead Mar 15 '15 at 13:29
  • $\begingroup$ @CliveNewstead: OK, thanks! I am with you that far - but why is $N$ not already small enough for that to be the case? $\endgroup$ – String Mar 15 '15 at 13:31
  • $\begingroup$ (Sorry, looks like I edited my comment as you were writing yours.) The reason is that the preimage of $U_i$ is a bunch of homeomorphic copies of $U_i$. One of these copies is $\widetilde{U}_i$, but there might be others; if $\widetilde{F}(N \times \{t_i\})$ contains elements not in $\widetilde{U}_i$ then you can't compose $p^{-1}$ because domains and codomains don't match up. $\endgroup$ – Clive Newstead Mar 15 '15 at 13:33
  • $\begingroup$ ...I'm going to turn these comments into an answer, bear with me. $\endgroup$ – Clive Newstead Mar 15 '15 at 13:33
2
$\begingroup$

You need $\widetilde{F}(N \times \{t_i\})$ to be contained in $\widetilde{U}_i$ so that you can use the homeomorphism $p$. The reason is that the preimage of $U_i$ is a bunch of homeomorphic copies of $U_i$. One of these copies is $\widetilde{U}_i$, which contains $\widetilde{F}(y_0,t_i)$, but there might be others; if $\widetilde{F}(N \times \{t_i\})$ contains elements not in $\widetilde{U}_i$ then you can't compose with $p^{-1}$ because domains and codomains don't match up.

$\endgroup$
  • $\begingroup$ Thank you very much! That clarified the whole thing. $\endgroup$ – String Mar 15 '15 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.