= Solution
{c}
\b[With <local path connectedness>, the intended conclusion holds.] Choose a <base point> $x_0$ and the standard <covering map>
$$
p:\mathbb R\longrightarrow S^1,\qquad p(t)=e^{2\pi it}.
$$
The <fundamental group> of the <circle> is $\mathbb Z$. Since the image of a <finite group> under a <group homomorphism> is finite and $\mathbb Z$ has no nontrivial finite <subgroups>,
$$
f_*:\pi_1(X,x_0)\longrightarrow\pi_1(S^1,f(x_0))\cong\mathbb Z
$$
is trivial. If $X$ is also <locally path-connected>, the <lifting criterion for a covering space> supplies a <continuous map> $\widetilde f:X\to\mathbb R$ with $p\widetilde f=f$, because $p_*\pi_1(\mathbb R)=0$. The <homotopy>
$$
\boxed{H(x,s)=p\big((1-s)\widetilde f(x)+s\widetilde f(x_0)\big)}
$$
runs from $f$ to the <constant function> $f(x_0)$. This proves the claimed <null-homotopy> under that additional hypothesis, in particular for <CW complexes> and the <topological graphs> considered later in the paper.
\b[For arbitrary <path-connected spaces>, the printed assertion is false.] Here is an explicit <Warsaw circle> counterexample, including the obstruction that the usual lifting argument misses. In the <Euclidean plane> take
$$
G=\{(x,\sin(\pi/x)):0<x\leq1\},\qquad
L=\{0\}\times[-1,1].
$$
Join $P=(1,0)$ to $Q=(0,-1)$ by the polygonal <arc> $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its <subspace topology>. The <arc> meets $G\cup L$ only at $P,Q$. The <topologist's sine curve> $G\cup L$ has two <path components>, but the added <arc> joins them, making $W$ <path-connected>.
Every <loop (topology)> in $W$ is contained in
$$
T_\delta=L\cup A\cup\{(x,\sin(\pi/x)):\delta\leq x\leq1\}
$$
for some $\delta>0$. To justify the assertion, a <path> cannot reach $L$ along the oscillating tail: the <intermediate value theorem> for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting <continuity>. The only alternative route from the tail to $L$ uses $P$ and the joining <arc>. If a <loop (topology)> visited tail points with $x\to0$, a convergent subsequence of their parameter values would approach a time at which the <loop (topology)> lies in $L$. On a sufficiently small parameter <interval> around that time its image avoids $P$. That would give a forbidden <path> from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_\delta$ is a finite <tree>, in fact a single <arc>, so every such <loop (topology)> contracts. Consequently
$$
\pi_1(W)=1.
$$
Now define a <continuous map> $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the <pasting lemma for closed subspaces> proves <continuity>.
There is no <continuous lift> $\widetilde h:W\to\mathbb R$. Any such lift is <integer>-valued on $G\cup L$, so it is constant on each of the <connected spaces> $G,L$. Since $G$ accumulates on $L$, <continuity> forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a <null-homotopy>, the <homotopy lifting theorem for a covering map>, starting with a lift of the final <constant function> and reversing time, would produce a <continuous lift> of $h$. This theorem needs no <local path connectedness> of the parameter space. Hence $h$ is not <null-homotopic>, although $W$ has finite, indeed trivial, <fundamental group>.
The native PDF explicitly assumes only <path connectedness> here. The missing <local path connectedness> cannot be inferred from the finiteness of the <fundamental group>.
Back to article page
