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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.