Hi, I’d greatly appreciate it if someone could illuminate the last three lines of this proof

Here is some context to aid in reading it:

[; C(I;Y) ;], where [; I ;] denotes the unit interval, is the space of all paths in [; Y ;], endowed with the topology generated by the following basis:

Let [; cup U*i ;], [; P = { t_0=0 ,t_1,…, t_n=1 } ;], be a collection of open sets where each [; U_i ;] comes from some basis for the topology on [; Y;] and [; P ;] is a partition on the unit interval [; I ;]. Furthermore the union is over the following set of indices: [; J = {1,2,…,n } ;]. (I wasn’t able to typeset a union no matter how hard I tried.. for whatever reason). Then the set [; A(t_0,t_1,..,t_n;U_1,U_2,..,U_n) ;] is the set of all paths [; a in C(I;Y) ;] such that [; a([t*{i-1},t_i]) in U_i ;] for all [; i ;]. We endow [; C(I;X) ;] with the same topology.

[; f : X rightarrow Y ;] a local homeomorphism of topological spaces has the “unique path lifting property”, if for all paths [; a in C(I;Y) ;] and [; x in X ;] such that [; f(x) = a(0) ;], there is exactly one lifted path [; tilde{a} in C(I;X) ;] such that [; tilde{a}(0) = x ;], the lift with respect to $f$.

I understand all of the proof besides the third to last line, since I see that [; tilde{b}([t_0,t_1]) = phi_1(b([t_0,t_1]) ;] as both lifts agree at [;t = t_0 = 0;], by construction since [;tilde{b}(0) = x’ in U_1;], so by unique path lifting they agree on [;[t_0,t_1];]. The issue I have is why this also holds for [;i=2,3,…,n;]

Thanks for reading !