Let $X$ be a topological space and let $\gamma, \delta : [0,1] \rightarrow X$ be two paths from $x$ to $y$. Now define $\widehat{\gamma}: [0,2] \rightarrow X$ by $$\widehat{\gamma}(t) = \begin{cases} \gamma(t) & \text{if $t \in [0,1]$,} \\ y & \text{if $t \in [1,2]$.} \end{cases}$$ So basically $\widehat{\gamma}$ does what $\gamma$ does and then stays at the end-point for another unit of time.

Similarly define $\widehat{\delta} : [0,2] \rightarrow X$. How can I show that if $\widehat{\gamma}$ and $\widehat{\delta}$ are path-homotopic, then $\gamma$ and $\delta$ are path-homotopic? (Of course a path-homotopy between $\widehat{\gamma}$ and $\widehat{\delta}$ would be a map $H: [0,1] \times [0,2] \rightarrow X$ with the usual conditions)

We can't just "trim" the given homotopy because the intermediate paths between $\widehat{\gamma}$ and $\widehat{\delta}$ need not pass through $y$ when $t=1$.

I am sure some people are able to draw a rectangle, cut it into pieces and "see" the required homotopy like an oracle but I am afraid I lack such skills.


Have you tried the homeomorphism $I\times I\to[0,2]\times I$ given by

$h(s,t)=\begin{cases} s(3t+1,\ 0)+(1-s)(0,t)& \text{ if }\ t\le\frac13\\ s(2,\ 3t-1)+(1-s)(0,t)& \text{ if }\ \frac13\le t\le\frac23\\ s(4-3t,\ 1)+(1-s)(0,t)& \text{ if }\ t\ge\frac23 \end{cases}$

Intuitively, this stretches the rectangle $I\times[\frac13,\frac23]$ to the trapezoid with vertices $(0,\frac13),\ (2,0),\ (2,1),\ (0,\frac23).$

Then compose with the given homotopy $H$. For $s\in\{0,1\}$ (the endpoints of the paths) $H\circ h(s,t)$ equals $x,y$ respectively, independent of the time $t$, so it is a path-homotopy from $\gamma$ to $\delta$.

The formula above can be generalized. Assume that $a:[0,p]\to X$ and $b:[0,q]\to X$ are paths of possibly different length $p,q\ge 0$. If we denote by $r_x$ (or just $r$) the constant path of length $r$ at the point $x$, then $r\cdot a$ is a path of length $r+p$ which coincides with $a$ on $[0,p]$ and is constant $a(p)$ during $[p,r+p]$. If $a$ and $b$ are paths that both start at $x=a(0)=b(0)$ and both end at $y=a(p)=b(q)$, then we say that $a,b$ are equivalent if there are $r,s\ge 0$ such that $$r\cdot p\simeq s\cdot q$$ The above is the special case $p,q,r,s=1$.

Let's assume that $a$ and $b$ are equivalent via $H:[0,r+p]×I\to X$. Then

$\varphi(x,t)=\begin{cases} x(3tr+p,\ 0)+(1-x)(0,t)& \text{ if }\ t\le\frac13\\ x(r+p,\ 3t-1)+(1-x)(0,t)& \text{ if }\ \frac13\le t\le\frac23\\ x(3(1-t)s+q,\ 1)+(1-x)(0,t)& \text{ if }\ t\ge\frac23 \end{cases}$
is a homeomorphism between $I×I$ and $[0,s+q]×I$. The composition $Hφ$ is then a homotopy between $a'(x)=a(px)$ and $b'(x)=b(qx)$, the "unifications" of $a,b$. Conversely, if $H'$ is a homotopy $a'\simeq b'$ between paths of length $1$, then $H'φ^{-1}$ gives a homotopy $r\cdot a\simeq s\cdot b$.
In particular, this shows that equivalent paths of same length are homotopic.

This also shows how the two definitions of the fundamental group(oid) that you find in literature (one using only paths of length $1$, the other allowing paths of arbitrary length) are equivalent.


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