... Fundamental group and Homotopy as Natural Transformation

Today, I got an idea about homotopy and fundamental group. So I am writing this Blog. I have not taken topology and homotopy seriously, so I hope you can point out some mistakes and add your point!

This blog post introduces how to use category theory concepts to explore fundamental ideas in topology, such as continuous functions, homotopy, and the fundamental group, along with their relationships. This approach aids in understanding the algebraic structures and topological invariants in topology.

Construct category from topology space.

Let us start with the point in $\mathrm{Top}$$\mathrm{Top}$. In $\mathrm{T}\mathrm{o}\mathrm{p}$$\rm Top$, the terminal object is a singleton set $\ast$$*$.

Consider a topology space $X$$X$, let $x\in X$$x\in X$ be a point. Then, we can replace it with the arrow(morphism) $\overline{x}:\ast \to X,\ast \mapsto x$$\bar x:*\to X,*\mapsto x$.

Thus, we can write $f(x)$$f(x)$ as $f\circ \overline{x}$$f\circ \bar x$, for convenience, $fx.$$fx.$

Let $X$$X$ be a topology space, then we can construct a category $\mathcal{X}$$\mathscr X$.

$\mathrm{Ob}(\mathcal{X}):\{x\in X|\ast \mapsto x\}$$\mathrm{Ob}(\mathscr X):\{x\in X|*\mapsto x\}$

$\mathrm{Mor}(\mathcal{X}):⟨p⟩:[0,1]\times \ast \to X,x\to x^{\prime }$$\mathrm{Mor}(\mathscr{X}):\langle p\rangle:[0,1]\times*\to X,x\to x'$ which is the equivalence class of path homotopy and $⟨p⟩(0)=x,⟨p⟩(1)=x^{\prime }$$\langle p\rangle(0)=x,\langle p\rangle(1)=x'$.

Remark

$†.$$\dagger.$ Observe that every path is a homotopy from $\overline{x}$$\bar x$ to ${\overline{x}}^{\prime }$$\bar x'$ !

$†.$$\dagger.$ The equivalence relation class of $x\sim y⟺\mathrm{\exists }p,p(0)=x,p(1)=y$$x\sim y\iff \exists p, p(0)=x,p(1)=y$ is each path connected component.

Thus in $\mathcal{X}$$\mathscr X$, evrery homomorphism is isomorphism(path is invertible). i.e. $\mathcal{X}$$\mathscr X$ is disjoint union of groupoids( each component is a groupid. )

Continuous function as a functor.

Let $f:X\to Y$$f:X\to Y$ be a continuous function.

Then we can induce a functor $f:\mathcal{X}\to \mathcal{Y}$$f:\mathscr X\to\mathscr Y$.

$f:x\mapsto y$$f:x\mapsto y$, $f(⟨p⟩):=⟨f\circ p⟩$$f(\langle p\rangle):=\langle f\circ p\rangle$ . $f(⟨a⟩\circ ⟨b⟩)=f(⟨a\circ b⟩)=f(⟨a⟩)\circ f(⟨b⟩)$$f(\langle a\rangle\circ\langle b\rangle)=f(\langle a\circ b\rangle)=f(\langle a\rangle)\circ f(\langle b\rangle)$.

Homotopy as natural transformation.

natural transformation in nLab (ncatlab.org)

Consider $f,g:X\to Y$$f,g:X\to Y$ and $h:[0,1]\times X\to Y,h(0,x)=fx,h(1,x)=gx$$h:[0,1]\times X\to Y,h(0,x)=fx,h(1,x)=gx$,

we can induce a natural transformation from the functor $h:(f:\mathcal{X}\to \mathcal{Y})⟹g:\mathcal{X}\to \mathcal{Y}$$h:(f:\mathscr X\to \mathscr Y)\Longrightarrow g:\mathscr X\to \mathscr Y$.

Indeed, natural isomorphism.

Homotopy of two spaces as Categorical equivalence .

We know that in $\mathrm{hTop}$$\mathrm{hTop}$ , the isomorphism of $X,Y$$X,Y$ becomes $f:X\to Y,g:Y\to X,\theta :f\circ g\simeq {\mathrm{id}}_Y,\psi :g\circ f\simeq {\mathrm{id}}_X$$f:X\to Y,g:Y\to X,\theta:f\circ g\simeq \mathrm{id}_Y,\psi :g\circ f\simeq\mathrm{id}_X$.

That is, the Categorical equivalence!

Fundamental Group

Consider an object $x_i\in \mathcal{X}$$x_i\in \mathscr X$, then the automorphism of $x$$x$ is the fundamental group ${\pi }_1(X,x_i)$$\pi_1(X,x_i)$ .

The fundamental group is not dependent on the choice of $x_i$$x_i$ if $X$$X$ is path-connected.

Since if $X$$X$ is path-connected, then for every $x,x^{\prime }\in \mathcal{X},x\cong x^{\prime }\Rightarrow {\pi }_1(X,x)\cong {\pi }_1(X,x^{\prime })$$x,x'\in\mathscr X,x\cong x'\Rightarrow\pi_1(X,x)\cong\pi_{1}(X,x')$.

In general, consider a locally small category $\mathcal{C}$$\mathcal C$, let $f:A\to B$$f: A\to B$ be the isomorphism, then ${\mathrm{Aut}}_{\mathcal{C}}(A)\cong {\mathrm{Aut}}_{\mathcal{C}}(B)$$\mathrm{Aut}_{\mathcal C}(A)\cong \mathrm{Aut}_{\mathcal C}(B)$.

Since $T_f:g⟼f^{-1}\circ g\circ f$$T_f:g\longmapsto f^{-1}\circ g\circ f$ is the group isomorphism.

Let $X,Y$$X,Y$ be two path-connected spaces with a base point, then $f:(X,x_0)\to (Y,y_0)$$f:(X,x_0)\to (Y,y_0)$ will induce a group homomorphism.

i.e. ${\pi }_1(f):{\pi }_1(X,x_0)\to {\pi }_1(Y,y_0),f⟨p⟩=⟨f\circ p⟩$$\pi_1 (f):\pi_1(X,x_0)\to\pi_1(Y,y_0),f\langle p\rangle=\langle f\circ p\rangle$. Since $f$$f$ is a functor, $f$$f$ is a group homomorphism.

It is obvious that the fundamental group is a functor from $\mathrm{T}\mathrm{o}{\mathrm{p}}_{\ast }\to \mathrm{G}\mathrm{r}\mathrm{p}$$\rm Top_*\to Grp$.

Moreover, we can see that it is a functor from $\mathrm{h}\mathrm{T}\mathrm{o}{\mathrm{p}}_{\ast }\to \mathrm{G}\mathrm{r}\mathrm{p}$$\rm hTop_*\to Grp$.

We can see it when we prove that the fundamental group of a path-connected space does not depend on the base point.

Lemma.

Let $f\simeq g$$f\simeq g$ be the continuous functions from $X$$X$ to $Y$$Y$, $h(t,x)$$h(t,x)$ be the homotopy between $f$$f$ and $g$$g$, then ${\pi }_1(f){\pi }_1(X,x_0)\cong {\pi }_1(g){\pi }_1(X,x_0)$$\pi_1(f)\pi_1(X,x_0)\cong\pi_1(g)\pi_1(X,x_0)$.

Proof.

View $h$$h$ as the natural isomorphism, $h(f)=g$$h(f)=g$, therefore $h(f\circ ⟨a⟩)=g\circ ⟨a⟩$$h(f\circ\langle a\rangle)=g\circ \langle a\rangle$. Since $h$$h$ is invertible, thus $h$$h$ induced a group isomorphism. $◻$$\square$

Or we might say that $h$$h$ is the natural isomorphism between ${\pi }_1(f)\simeq {\pi }_1(g)$$\pi_1(f)\simeq\pi_1(g)$.

Fundamental Group is homotopy invariance.

For convenience, denote $G_x,H_y$$G_x,H_y$ be the fundamental group.

Proof

${\pi }_1(f\circ g)(H_y)={\pi }_1(f)\circ {\pi }_1(g)(H_y)\cong {\pi }_1({\mathrm{id}}_Y)(H_y)=H_y$$\pi_1(f\circ g)(H_y)=\pi_1(f)\circ\pi_1(g)(H_y)\cong\pi_1(\mathrm{id}_Y)(H_y)=H_y$

${\pi }_1(g\circ f)G_x={\pi }_1(g)\circ {\pi }_1(f)G_x\cong {\pi }_1({\mathrm{id}}_{\mathrm{X}})G_x=G_x$$\pi_1(g\circ f)G_x=\pi_1(g)\circ\pi_1(f)G_x\cong\pi_1(\mathrm{id_X})G_x=G_x$

Thus ${\pi }_1(f),{\pi }_1(g)$$\pi_1(f),\pi_1(g)$ are group isomorphism. Thus $G_x\cong H_y.◻$$G_x\cong H_y. \square$