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Algebraic topology translates questions about topological spaces and continuous maps into questions about groups and other structures from algebra. A topological invariant compresses the enormous amount of data in a space into something one can calculate. For example, the fundamental group records which based loops can be continuously deformed into one another. A filled disk has trivial fundamental group, whereas a circle has fundamental group $\mathbb{Z}$: an integer remembers how many times a loop goes around.

This compression also respects maps. By functoriality of the fundamental group, a continuous map preserving the chosen basepoints induces a group homomorphism; successive maps induce successive group homomorphisms. Thus geometry supplies arithmetic constraints. A proposed homotopy equivalence must induce an isomorphism, so incompatible fundamental groups rule it out before anyone attempts to construct the deformation.

The beauty is that a whole annulus can retain exactly the same homotopy type as its inner circle. Its thickness disappears under a deformation retraction, while the obstruction to shrinking a loop survives. Equal fundamental groups alone do not establish a homeomorphism or a homotopy equivalence between arbitrary topological spaces.

This turns a visual question into a reusable test. Suppose a continuous map sends the inner circle of an annulus once around another circle. Could it extend over a filled disk spanning that boundary? Such an extension would supply a homotopy shrinking the image loop to a point. Its winding number would have to change from one to zero, which homotopy invariance of winding number forbids. Algebra rules out the extension even when no particular formula for it has been proposed.

A curve $\gamma :[0,1]\to X$ whose endpoints agree, $\gamma (0)=\gamma (1)$. This is the same unbased notion as a loop; self-intersections are allowed.

A deformation retraction of $X$ onto a subspace $A$ is a homotopy $H:X\times [0,1]\to X$ such that $H(x,0)=x$, $H(x,1)=r(x)\in A$, and $H(a,t)=a$ for every $a\in A$ and every $t$. Thus the retract is fixed pointwise throughout. With inclusion $i:A\hookrightarrow X$, the equations $r\circ i={\mathrm{id}}_A$ and $i\circ r\simeq {\mathrm{id}}_X$ exhibit a homotopy equivalence. See Hatcher, Algebraic Topology, Chapter 0, pp. 2–3.

Take the closed annulusAn exact deformation retraction onto its inner circle isThe denominator is nonzero everywhere on the annulus. If the starting radius is $\rho =\Vert x\Vert$, its radius at time $t$ is $(1-t)\rho +t$. This lies between $1$ and $\rho \le 2$, verifying that the entire homotopy stays in the original annulus. When $\rho =1$, the point is stationary, as required.

For $x=(0,2)$, times $0,1/2,1$ give $(0,2),(0,3/2),(0,1)$. A radius-$3/2$ loop similarly becomes a radius-$5/4$ loop halfway through, then the unit circle. Its angle and winding number are preserved. This is useful because computing the fundamental group of the annulus reduces to computing that of the circle; the explicit homotopy equivalence establishes why the answer transfers.

A homotopy between continuous maps $f,g:X\to Y$ is a continuous map $H:X\times [0,1]\to Y$ satisfying $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Think of $t$ as time: continuity is required jointly in position and time. The target $Y$ specifies where every intermediate point is allowed to go. See Hatcher, Algebraic Topology, Chapter 0, pp. 3–4.

For the filled disk $D^2=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le 1\}$, the formulais a contraction of a topological space: it starts at the identity map and ends at the constant map to zero. The Euclidean norm obeys $\Vert H(x,t)\Vert =(1-t)\Vert x\Vert \le 1$, so the formula never leaves the disk. For $x=(3/5,4/5)$, the midpoint of the deformation is $(3/10,2/5)$, whose Euclidean norm is $1/2$.

On the circle $S^1=\{x:\Vert x\Vert =1\}$, the same formula fails: at $t=1/2$ every image has Euclidean norm $1/2$, so lies outside $S^1$. This checks one failed candidate; the nonzero winding number of a once-around loop proves that no contraction within the circle exists. The distinction is between allowing the filling and requiring the deformation to stay on its boundary.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

Let $C\subset {\mathbb{P}}_{\mathbb{C}}^r$, $r\ge 2$, be a smooth irreducible nondegenerate projective variety of dimension one and degree $d\ge 2$. Let $U$ be the open subset of the dual projective space consisting of hyperplanes meeting $C$ in $d$ distinct points. The incidence projectionis a $d$-sheeted covering space in the complex analytic topology. Its monodromy is the full symmetric group $S_d$.^[ref]

The incidence space choosing two distinct ordered intersection points is irreducible, so the monodromy action of a covering space is two-transitive. A small loop around an ordinary tangent hyperplane exchanges its two nearby intersection points and fixes the others. Conjugating this transposition by double transitivity gives every transposition, and these generate $S_d$. In particular the action on $k$-element subsets is transitive for each $0\le k\le d$. For a degree-one projective line, a transverse section has one point and the corresponding group is the trivial $S_1$.

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.

Use a topological graph consisting of the ray $[0,\infty )$ with vertices $v_n=n$ and one oriented graph loop ${\ell }_n$ attached at each $v_n$, for $n=0,1,2,\dots$. Set the base point to $b=v_0$. This is a connected, locally finite graph; it is deliberately infinite.

Define continuous maps $f,g:X\to X$ byHere each graph loop is parametrized by $s\in [0,1]$ with its endpoints at the attaching vertex. The formulas agree at every vertex, and their restrictions to every closed edge are continuous. The weak topology of a CW complex therefore makes $f,g$ continuous maps. Directly on the ray and on each graph loop,

We check the two fundamental group maps at exactly the specified base points. The ray is a spanning tree. At $b=v_0$ a free basis of a group consists of $a_n$, obtained by going along the ray to $v_n$, following ${\ell }_n$, and returning. At $f(b)=v_1$, use the corresponding free basis of a group $c_n$ with connecting paths starting at $v_1$; in particular $c_0$ travels backwards along the first edge before following ${\ell }_0$. The standard spanning tree description applies to this infinite topological graph as well: every loop and every homotopy between based loops has compact image, so lies in a finite subcomplex.

The induced group homomorphisms areThus $f_{\ast }$ misses $c_0$: words in $c_1,c_2,\dots$ cannot equal $c_0$ by the normal form for free groups and free product of groups. Consequently $f_{\ast }$ is not surjective. Meanwhile $c_0$ is a nontrivial free group element killed by $g_{\ast }$, so $g_{\ast }$ is not injective. Neither induced map is a group isomorphism, while functoriality of the fundamental group correctly gives $g_{\ast }{f}_{\ast }=\mathrm{id}$.

Let $X=U\cup V$ be an open cover such that $U$, $V$ and $U\cap V$ are nonempty and path-connected, and choose a base point $x_0\in U\cap V$. The inclusions induce a pushout of groups:The maps from the overlap need not be injective. The amalgamated free product here means the universal group receiving the two inclusion-induced homomorphisms with their overlap relations.

A loop can be subdivided into pieces in $U$ or $V$; connecting their endpoints through the path-connected overlap expresses its class using the two local fundamental groups. Subdividing a homotopy shows that the overlap relations are exactly the identifications needed.

The overlap condition matters. Two open arcs can cover a circle while their intersection has two components. Both arcs have trivial fundamental group, but the circle has fundamental group $\mathbb{Z}$. The displayed single-basepoint formula cannot be applied to that disconnected overlap; a groupoid version keeps the extra component information.