# Homotopy of Paths

For this post we will explain what is a homotopy of paths.

Source: Topology (2nd Economy Edition)   The book above is a nice introductory book on Topology, which includes a section of introductory Algebraic Topology.

Definition: If f and f’ are continuous maps of the space X into the space Y, we say that f is homotopic to f’ if there is a continuous F: X x I -> Y such that

F(x, 0)=f(x) and F(x,1) = f'(x)

for each x. The map F is called a homotopy between f and f’. If f is homotopic to f’, we write $f \simeq f'$.

If f and f’ are two paths in X, there is a stronger relation, called path homotopy, which requires that the end points of the path remain fixed during the deformation. We write $f \simeq_p f'$ if f and f’ are path homotopic.

Next, we will prove that the relations $\simeq$ and $\simeq_p$ are equivalence relations.

If f is a path, we shall denote its path-homotopy equivalence class by [f].

Proof: We shall verify the properties of an equivalence relation, namely reflexivity, symmetry and transitivity.

Reflexivity:

Given f, it is rather easy to see that $f \simeq f$. The map F(x,t) is the required homotopy.

F(x,0)=f(x) and F(x,1)=f(x) is clearly satisfied.

If f is a path, then F is certainly a path homotopy, since f and f itself has the same initial point and final point.

Symmetry:

Next we shall show that given $f \simeq f'$, we have $f' \simeq f$. Let F be a homotopy between f and f’. We can then verify that G(x,t) = F(x, 1-t) is a homotopy between f’ and f.

G(x,0) = F(x, 1)=f’ (x)

G(x,1) = F(x, 0) = f(x)

Furthermore, if F is a path homotopy, so is G.

G(0,t)=F(0, 1-t) = $x_0$

G(1,t)=F(1,1-t) = $x_1$

Transitivity:

Next, suppose that $f \simeq f'$ and $f' \simeq f''$, we show that $f \simeq f''$. Let F be a homotopy between f and f’, and let F’ be a homotopy between f’ and f”. This time, we need to define a slightly more complicated homotopy G: X x I -> Y by the equation $G(x,t) = \begin{cases} F(x,2t) &\text{for }t\in [0,\frac{1}{2}],\\ F'(x, 2t-1) &\text{for } t\in [\frac{1}{2}, 1].\end{cases}$

First, we need to check if the map G is well defined at t=1/2. When t=1/2, we have F(x,2t) = F(x,1)=f'(x) = F'(x,2t-1).

Because G is continuous on the two closed subsets X x [0, 1/2] and X x [1/2, 1] of XxI, it is continuous on all of X x I, by the pasting lemma.

Thus, we may see that G is the required homotopy between f and f”.

G(x,0)=F(x,0) = f(x)

G(x,1) = F’ (x, 1) = f”(x)

We can also check that if F and F’ are path homotopies, so is G.

G(0,t) = F(0, 2t) = $x_0$

G(1, t)=F'(1, 2t-1) = $x_1$ ## Author: mathtuition88

https://mathtuition88.com/

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