Fundamental group of $G/H$ without homotopy lifting/covering maps I am working on Exercise 3.24 of Rotman's algebraic topology text. It has been discussed several times on math.SE already (Discrete Normal Subgroup of a Simply Connected Topological Group. and Discrete Closed Subgroup H of a Simply Connected Topological Group G isomorphic to fundamental group of G / H., among others), but all of them use universal coverings or the homotopy lifting axiom, neither of which I've learned yet.
The problem is to show that if $G$ is a simply connected topological group, and if $H$ is a discrete closed normal subgroup, then $\pi_1(G/H,1)\cong H$.
I have worked on this for a while now and all I have left is to show that I can lift $f:[0,1]\to G/H$ into a function $\tilde f:[0,1]\to G$, but I'm not sure how to prove this lifting. In particular, what I'd like to show is the following statement: For every $g_0\in H$, given a loop $f:[0,1]\to G/H$ with $f(0)=f(1)=1$, there is a function $\tilde f:[0,1]\to G$ with $\tilde f(0)=g_0$ such that $\tilde f(t)H=f(t)$.
I think I'm supposed to use some open neighborhood $U$ of $1$ such that the family of $hU$'s, where $h\in H$, is disjoint. I can show that $U$ exists, and I think that the condition on $U$ is supposed to make the natural map $v:G\to G/H$ a homeomorphism on $U$, but I can't prove that this is true.
If somebody can give me a hint on how to continue, that'd be great. I know that this is related to covering spaces, etc., but I'd like to try solving this without using those ideas, if possible.
Also, in the book, the hint is to generalize a proof (it says a proof of 3.16, but if you're looking at the book, I'm actually trying to generalize Lemma 3.14). It's a bit long to reproduce here, so if somebody just can explain what $v(U)$ looks like (is it all of $G/H$? I think it might not be, but I'm not sure), that'd be good enough for now.
 A: You can prove the homotopy lifting lemma in this special case, not sure if that is what you are looking for.
So you know that your neighbourhood $U\ni 1$ exists.  Let's write $V=v(U)$.  There is nothing special about $1$ by translating $U$.  Also, we may assume $U$ is connected.
We know $v\vert U$ is continuous (restricting a continuous), open ($v$ being open map and $U$ is open) and bijective $U\to V$, so is a homeomorphism.
Digression: $V$ need not be the whole of $G/H$.  For example, take $G=\mathbb{R}$ and $H=2\pi\mathbb{Z}$, $G\to G/H\cong S^1$ by $t\mapsto e^{it}$.  You can't have the whole of $S^1$ homeomorphic to an open subset of $\mathbb{R}$ since there are no open nonempty compact subsets of $\mathbb{R}$.

Now $V\bar{g}$ is open for all $\bar{g}\in G/H$ and they form an open cover of $G/H$.  So
$$
\{f^{-1}(V\bar{g})\mid g\in G, \bar{g}:=gH\}
$$
forms an open cover of $[0,1]$ by open intervals.  As $[0,1]$ is connected compact, there is a finite subcover of open overlapping subintervals.
Now we lift $f$ to $\tilde{f}$ on each: Starting with $0\in [0,1]$, $0=:t_1\in I_1\subseteq f^{-1}(V\overline{g_1})$.  Then $v^{-1}V\overline{g_1}$ is a disjoint union of translates of $U$, so select the one that contains $g_0$.  For every $t\in I_1$, we select $\tilde{f}(t)$ to be the unique element in this translate such that $v(\tilde{f}(t))=f(t)$ (i.e., $\tilde{f}(t)H=f(t)$).  Since the intervals are overlapping, select $I_2\subseteq f^{-1}(V\overline{g_2})$, $I_2\cap I_1\neq\varnothing$, $t_2\in I_1\cap I_2$ and continue.  You can check $\tilde{f}\colon[0,1]\to G$ is well-defined.
