Fundamental Group

Definition

If the starting and ending point of a path are the same, then this path is called a loop The common starting and ending point $x_0$ is referred to as the basepoint

The set of all homotopy classes $[f]$ of loops $f:I\to X$ at the basepoint $x_0$ is denoted ${\pi }_1(X,x_0)$

Proposition 1.3 ( Algebraic Topology, p.35)

${\pi }_1(X,x_0)$ is a group with respect to the product $[f][g]=[f\cdot g]$

This group is called the fundamental group of $X$ at the basepoint $x_0$

Further we can show that the map ${\beta }_h:{\pi }_1(X,x_1)\to {\pi }_1(X,x_0)$ is an isomorphism

Thus if $X$ is path-connected, the group ${\pi }_1(X,x_0)$ is, up to isomorphism, independent of the choice of basepoint $x_0$. In this case the notation ${\pi }_1(X,x_0)$ is often abbreviated to ${\pi }_1(X)$ or ${\pi }_{1}X$.

Functoriality

We can view ${\pi }_1$ as a covariant functor $To{p}_{\ast }\to Grp$ (a covariant functor preserves the direction of morphisms)

notation

We use ${\phi }_{\ast }$ to denote the translated $\phi$ after applying the functor

• For objects: $(X,x_0)\mapsto {\pi }_1(X,x_0)$
• For morphisms: $\phi \mapsto {\phi }_{\ast }$ where
  • $\phi :(X,x_0)\to (Y,y_0)$ is a continuous map
  • ${\phi }_{\ast }:{\pi }_1(X,x_0)\to {\pi }_1(Y,y_0)$ is a group homomorphism with ${\phi }_{\ast }[f]=[\phi f]$
  • identity preservation: The continuous map $i{d}_X:X\to X$ induces the identity homomorphism $i{d}_{{\pi }_1(X,x_0)}:{\pi }_1(X,x_0)\to {\pi }_1(X,x_0)$
  • composition map: $(\phi \psi {)}_{\ast }={\phi }_{\ast }{\psi }_{\ast }$ for composition $(X,x_0)\stackrel{\psi }{\to }(Y,y_0)\stackrel{\phi }{\to }(Z,z_0)$
    • proof: For any $[f]\in {\pi }_1(X,x_0)$ we have $(\phi \psi {)}_{\ast }([f])=[\phi \psi f]={\phi }_{\ast }[\psi f]={\phi }_{\ast }{\psi }_{\ast }[f]$

Note that any continuous map $\phi :X\to Y$ induces a group homomorphism ${\phi }_{\ast }:{\pi }_1(X,x_0)\to {\pi }_1(Y,\phi (x_0))$ given by ${\phi }_{\ast }[f]=[\phi f]$ ($f$ is a loop here)

We can check ${\phi }_{\ast }$ is indeed a homomorphism: Since if $f$ and $g$ are loops at $x_0$, then $\phi (f\cdot g)=(\phi f)\cdot (\phi g)$ as loops at $y_0$ (see 13.1 Definition)

Theorem 1.18/2.26

Algebra Topology Notes, page 19

If $\phi :X\to Y$ is a homotopy equivalence, then ${\phi }_{\ast }:{\pi }_1(X,x_0)\to {\pi }_1(Y,\phi (x_0))$ is an isomorphism.

geometric examples

an infinite cyclic group (free group on one generator)

AB are not linked nonabelian free group on two generators

AB are linked free abelian group on two generators

Important

One can often show that two spaces are not homeomorphic by showing that their fundamental groups are not isomorphic,

Check this nice visualization video on YouTube

Example: Circle

We show that ${\pi }_1(S_1)\approx \mathbb{Z}$ ($\approx$ meaning isomorphism here) Claim

Theorem 1.7 Algebraic Topology, p.38

${\pi }_1(S^1)$ is an infinite cyclic group generated by the homotopy class of the loop $\omega (s)=(\cos 2\pi s,\sin 2\pi s)$ based at $(1,0)$

Using the definition of Covering Space: …

• Proof

Proposition 1.12

Algebraic Topology, page 43

${\pi }_1(X\times Y)$ is isomorphic to ${\pi }_1(X)\times {\pi }_1(Y)$ if $X$ and $Y$ are path-connected

Proposition 1.14

Algebraic Topology, page 44

${\pi }_1(S^n)=0$ if $n\ge 2$

Corollary 1.16

Algebraic Topology, page 45

${\mathbb{R}}^2$ is not homeomorphic to ${\mathbb{R}}^n$ for $n\ne 2$

For $n=1:$ Suppose there is a homeomorphism $f$ between ${\mathbb{R}}^2$ and ${\mathbb{R}}^1$ However, we can see that ${\mathbb{R}}^2-\{0\}$ is path-connected while ${\mathbb{R}}^1-\{f(0)\}=(-\infty ,f(0))\cup (f(0),\infty )$ is not. This implies that they cannot be homeomorphic.

For $n>2:$ Suppose there is a homeomorphism $f$ between ${\mathbb{R}}^2$ and ${\mathbb{R}}^n$, then ${\mathbb{R}}^2-\{0\}$ must still have a homeomorphism to ${\mathbb{R}}^2-\{f(0)\}$ (since $f$ is bijective) We know ${\mathbb{R}}^n-\{0\}$ is homeomorphic to $S^{n-1}\times \mathbb{R}$ $⟹$ ${\pi }_1({\mathbb{R}}^n-\{x\})$ is isomorphic to ${\pi }_1(S^{n-1})\times {\pi }_1(\mathbb{R})={\pi }_1(S^{n-1})$ (by Proposition 1.14) So ${\pi }_1({\mathbb{R}}^n-\{x\})=\mathbb{Z}$ for $n=2$ and ${\pi }_1({\mathbb{R}}^n-\{x\})=0$ otherwise. Therefore, such a homeomorphism cannot exist. (by Theorem 1.18/2.26, note that homeomorphism implies homotopy equivalence)

Fundamental Theorem of Algebra

Algebraic Topology, page 40

Every nonconstant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$

Assume $p(z)=z^n+a_1{z}^{n-1}+\cdots +a_n$ has no roots in $\mathbb{C}$. For each radius $r\ge 0$, consider the loop obtained by mapping the circle $|z|=r$ through $p$ and then normalizing to the unit circle:

$$f_r(s)=\frac{p(r{e}^{2\pi is})/p(r)}{\left|p(r{e}^{2\pi is})/p(r)\right|}.$$

Because $p$ has no zeros, this is well-defined. As $r$ varies, the loops $f_r$ form a based homotopy. At $r=0$, the loop is constant, so

$$[f_r]=0\in {\pi }_1(S^1)$$

for every $r$.

Now choose $r$ sufficiently large. On $|z|=r$, the leading term $z^n$ dominates all lower-order terms $|z^n|>|a_1{z}^{n-1}+\cdots +a_n|.$ Therefore the family

$$p_t(z)=z^n+t(a_1{z}^{n-1}+\cdots +a_n),0\le t\le 1,$$

has no zeros on $|z|=r$. This gives a homotopy between the loop induced by $p$ and the loop induced by $z^n$.

But the loop induced by $z^n$ is

$${\omega }_n(s)=e^{2\pi ins},$$

which winds around $S^1$ exactly $n$ times. Hence it represents

$$n\in {\pi }_1(S^1)\cong \mathbb{Z}.$$

Thus $n=[{\omega }_n]=[f_r]=0.$ So $n=0$, meaning $p$ is constant. Therefore every nonconstant complex polynomial must have a root in $\mathbb{C}$.

See the proof of this theorem using analysis: Fundamental Theorem of Algebra