View pass as a homotopy

I learn this from Topology: a Categorical View

Consider a pass $\gamma :[0,1]\to (X,\tau )$$\gamma:[0,1]\to (X,\tau)$,

$\gamma (0)=a,\gamma (1)=b$$\gamma(0)=a,\gamma(1)=b$

We can view a pass as a homotopy.

Review the definition of homotopy

Consider two continuous functions $f,g:=(X,{\tau }_1)\to (Y,{\tau }_2)$$f,g:=(X,\tau_1)\to (Y,\tau_2)$

A homotopy is a continuous function $h(x,t):(X,{\tau }_1)\times [0,1]\to (Y,{\tau }_2)$$h(x,t):(X,\tau_1)\times[0,1]\to (Y,\tau_2)$

$h(x,0)=f(x),h(x,1)=g(x)$$h(x,0)=f(x),h(x,1)=g(x)$

For example, Consider $f,g:=\mathbb{R}\to \mathbb{R}$$f,g:=\mathbb R\to \mathbb R$

$h(x,t)=tg(x)+(1-t)f(x)$$h(x,t)=tg(x)+(1-t)f(x)$ is a homotopy

Review the ''point'' in Category Theory is the map from the final object to an object

For example, in $\mathrm{S}\mathrm{e}\mathrm{t}$$\rm Set$, an element $x\in X$$x\in X$ can be viewed as $x:\ast \to x$$x:*\to x$

Thus $f(x)$$f(x)$ can be written as $f\circ x,fx$$f\circ x,fx$

Now consider the final object in $\mathrm{T}\mathrm{o}\mathrm{p}$$\rm Top$, $a,b$$a,b$ can be viewed as continuous functions $\ast \to a,\ast \to b$$*\to a,*\to b$

Thus $\gamma :[0,1]\to (X,\tau )$$\gamma:[0,1]\to (X,\tau)$ can be viewed as $h(x,t):\ast \times [0,1]\to (X,\tau )$$h(x,t):*\times[0,1]\to (X,\tau)$

$h(x,0)=a,h(x,1)=b$$h(x,0)=a,h(x,1)=b$