# Understanding a proof of lifting $F:Y\times I\rightarrow X$ to $\widetilde F:Y\times I\rightarrow \widetilde X$

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?

• 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 $U_i$. – Clive Newstead Mar 15 '15 at 13:29
• @CliveNewstead: OK, thanks! I am with you that far - but why is $N$ not already small enough for that to be the case? – String Mar 15 '15 at 13:31
• (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. – Clive Newstead Mar 15 '15 at 13:33
• ...I'm going to turn these comments into an answer, bear with me. – Clive Newstead Mar 15 '15 at 13:33

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.