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The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

For a covering map $p:X\to Y$ and base point $y$, the path lifting theorem lifts each based loop from each point of the fibre $p^{-1}(y)$. Its endpoint defines a permutation of that fibre; path reversal gives the inverse. The homotopy lifting theorem for a covering map makes the permutation depend only on the based homotopy class. These permutations form the monodromy group, the image of the fundamental group actionThe image is independent of the convention for composing routes; one may reverse the loop classes to obtain a left group action when route multiplication means following the first route and then the second.

For $p(w)=w^2$ on the punctured plane, the fibre over $1$ is $\{1,-1\}$, and a once-around based loop exchanges its two points. The monodromy group has two elements: only the parity of the winding number matters. For the covering map $p(w)=e^w$ from the complex plane to the punctured plane, the fibre over $1$ is $2\pi i\mathbb{Z}$, and a positive turn acts by $2\pi ik\mapsto 2\pi i(k+1)$. Its monodromy group is an infinite cyclic group. These are the geometric versions of logarithm and square-root continuation along different paths: the base point returns, while its lifted endpoint can move to another sheet.

Use the homotopy lifting theorem for a covering map: a homotopy of paths has a unique lift once its initial path is lifted. Also use uniqueness of lifts to a covering space: a lift of a constant path with a specified starting point is constant.

Let $\tilde{\gamma }$ be a loop based at ${\tilde{x}}_0$ whose image under $p_{\ast }$ is trivial. There is a homotopy $F:I\times I\to X$ from $p\tilde{\gamma }$ to the constant loop at $x_0$, fixing both endpoints. Lift this homotopy with initial path $\tilde{\gamma }$. Both vertical boundary paths are constant lifts at ${\tilde{x}}_0$. Its final path is a lift of the constant loop and hence is also constant at ${\tilde{x}}_0$. Thus the lifted homotopy contracts $\tilde{\gamma }$ relative to its endpoints. The kernel of the induced group homomorphism is trivial, provingNo universal-cover assumption is needed for this injection; it holds for any actual covering map.

Take a homotopy equivalence $f:X\to Y$ with inverse $g$, and universal covering maps $p_X,p_Y$. Since the covering spaces are simply connected, the path lifting theorem constructs lifts $\tilde{f}$ of $f{p}_X$ and $\tilde{g}$ of $g{p}_Y$, after specifying one value in the appropriate fiber. Explicitly, define the lift at a point by lifting the image of a path from the chosen base point; independence from the path follows from simple connectedness and homotopy lifting.

Lift a homotopy $gf\simeq 1_X$ starting at $\tilde{g}\tilde{f}$. The homotopy lifting theorem for a covering map makes its endpoint a map $d_X:\tilde{X}\to \tilde{X}$ with $p_X{d}_X=p_X$. Such a lift of the identity is a deck transformation and is a homeomorphism: lift the identity again with the base point image reversed, and uniqueness of path lifts makes the two compositions identities. Likewise $\tilde{f}\tilde{g}\simeq d_Y$ for a deck transformation of $\tilde{Y}$.

Thus both compositions are homotopy equivalences, even though the lifted homotopies need not end at the particular identity lift. Part (i) then provesKeeping the possible deck transformations is essential when the original homotopies do not fix base points.

A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\tilde{f}=f$. The covering homotopy theorem says that a homotopy $H:Y\times I\to X$ has a unique lift once a continuous lift of its initial map $H(-,0)$ is prescribed. No additional local connectivity assumption on $Y$ is needed for this homotopy-lifting statement.

To prove injectivity on fundamental groups, let a loop $\tilde{\gamma }$ based at $\tilde{b}$ project to a based null-homotopic loop. Lift its null-homotopy, starting with $\tilde{\gamma }$. The two endpoint tracks lift constant paths at $b$ starting at $\tilde{b}$, so remain at $\tilde{b}$ by uniqueness of path lifting. The final path lies in the discrete fibre over $b$ and starts at $\tilde{b}$, so it is constant. Thus the lifted homotopy contracts $\tilde{\gamma }$ relative its basepoint, provingIn the following constructions, take $x$ around the circle factor and $y$ around the projective-plane factor. Above each sheet point, the circle has one incoming and one outgoing lifted $x$ edge. A fixed sheet for $y$ carries an identity cover of ${\mathbb{RP}}^2$; a transposed pair of sheets is joined as the antipodal fibre of a sphere covering ${\mathbb{RP}}^2$. These pieces give genuine covering neighbourhoods at the wedge point as well as away from it.

With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe 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,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns 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.

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 takeJoin $P=(1,0)$ 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 in $W$ is contained infor 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 visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop 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 contracts. ConsequentlyNow 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 $\tilde{h}:W\to \mathbb{R}$. 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.

The Warsaw circle joins the two path components of a closed topologist's sine curve by an arc attached at their outer endpoints. It is path-connected and has trivial fundamental group, yet admits a continuous map to the circle with no continuous lift to its real-line covering space. Set the map equal to $1$ on the closed topologist's sine curve and wind once on the joining arc. A lift would be constant on each path component of the sine curve, with the same constant on both by continuity, but would change by $1$ along the joining arc. This proves that the map is not null-homotopic, by the homotopy lifting theorem for a covering map. Thus finiteness of the fundamental group alone cannot replace local path connectedness in the usual lifting criterion for a covering space.