Continuity of Homotopy of paths

Question

I've been trying to show that if any loop in a region is homotopic to a point, then any two paths are homotopic to one another. I've put an attempt at a solution below, but I cannot quite figure out how to verify continuity when $t=1/2$. If anyone can offer any suggestions I would be really appreciative.

Attempt:

Let $\Omega$ be a space that is simply-connected in the sense of loops. Let $\alpha, \beta$ be distinct points in $\Omega$. By assumption, $\Omega$ is also path-connected, and so there are paths $\gamma_1,\gamma_1 \colon I \to \Omega$, where $I = [0,1]$, that connect $\alpha$ to $\beta$. Define the loop $\gamma \colon I \to \Omega$, a loop with base point $\alpha$ by $$ \gamma(s) = \begin{cases} \gamma_1(2s) &\text{if } s \in [0,1/2]\\ \gamma_2(2-2s) &\text{ if } s \in [1/2,1]. \end{cases} $$ The space $\Omega$ is simply connected and so there is a point $z \in \Omega$ and a family of loop $\{\sigma_s(t)\}$ that continuous deforms $\gamma$ onto the point $z$. We want to find a family that takes the path $\gamma_1$ to $\gamma_2$ without moving the points $\alpha$ or $\beta$. Here is my suggested family $\{\tau_t(s)\}$, which I define in two pieces. When $0 \leq t < 1/2$, define $$ \tau_t(s) = \begin{cases} \sigma_s(0) &\text{if } s \in [0,t]\\ \sigma_t\left(\frac{s-t}{2(1-2t)}\right) &\text{ if } s \in [t,1-t]\\ \sigma_{1-s}(1/2) &\text{ if } s \in [1-t,1],\\ \end{cases} $$ for $t \in (1/2,1]$, we have $$ \tau_t(s) = \begin{cases} \sigma_s(0) &\text{if } s \in [0,1-t]\\ \sigma_t\left(\frac{(1-t)-s}{2(1-2t)}\right) &\text{ if } s \in [1-t,t]\\ \sigma_{1-s}(0) &\text{ if } s \in [t,1],\\ \end{cases} $$ and finally, when $t = 1/2$, we define $$ \tau_{1/2}(s) = \begin{cases} \sigma_{2s}(0) &\text{if } s \in [0,1/2]\\ \sigma_{2(1-s)}(0) &\text{ if } s \in [1/2,1].\\ \end{cases} $$

Answer 1

Have you seen basic constructions of path concatenations/inverses already? If so, I think you're making this more difficult than it needs to be. Path concatenation is always well-defined, and we will write this as $\gamma\cdot\eta$ for paths $\gamma,\eta$ with $\gamma(1)=\eta(0)$. We will also write $\bar\gamma$ for the "inverse path" of $\gamma$, i.e. $\bar\gamma(t)=\gamma(1-t)$.

Furthermore, recall that path concatenation is well-defined with respect to homotopy, i.e. if $[\gamma]$ denotes the homotopy class of $\gamma$, then defining $[\gamma]\cdot[\eta]=[\gamma\cdot\eta]$ is well-defined.

Now, if $\gamma$ and $\eta$ are two paths from $\alpha$ to $\beta$, the concatenation $\gamma\cdot\bar\eta$ is a loop based at $\alpha$, so it is homotopic to the constant path $c_{\alpha}$, i.e. we have $[c_{\alpha}]=[\gamma\cdot\bar\eta]=[\gamma]\cdot[\bar\eta]$. Concatenating on the right hand side by $[\eta]$ gives us

$$[\gamma]=[c_{\alpha}]\cdot[\eta]=[c_{\alpha}\cdot\eta]=[\eta],$$

since $c_{\alpha}\cdot\eta$ is homotopic to $\eta$. This precisely says then that $\gamma$ and $\eta$ are homotopic paths.

Answer 2

Let $\Omega$ be a space that is simply-connected in the sense of loops. Let $\alpha, \beta$ be distinct points in $\Omega$. By assumption, $\Omega$ is also path-connected, and so there are paths $\gamma_1,\gamma_1 \colon I \to \Omega$, where $I = [0,1]$, that connect $\alpha$ to $\beta$. Define the loop $\gamma \colon I \to \Omega$, a loop with base point $\alpha$ by $$ \gamma(s) = \begin{cases} \gamma_1(2s) &\text{if } s \in [0,1/2]\\ \gamma_2(2-2s) &\text{ if } s \in [1/2,1]. \end{cases} $$ The space $\Omega$ is simply connected and so there is a point $z \in \Omega$ and a family of loop $\{\sigma_s(t)\}$ that continuous deforms $\gamma$ onto the point $z$. We want to find a family that takes the path $\gamma_1$ to $\gamma_2$ without moving the points $\alpha$ or $\beta$. Here is my suggested family $\{\tau_t(s)\}$, which I define in two pieces. When $0 \leq t < 1/2$, define $$ \tau_t(s) = \begin{cases} \sigma_s(0) &\text{if } s \in [0,t]\\ \sigma_t\left(\frac{s-t}{2(1-2t)}\right) &\text{ if } s \in [t,1-t]\\ \sigma_{1-s}(1/2) &\text{ if } s \in [1-t,1],\\ \end{cases} $$

when $t = 1/2$, we define $$ \tau_{1/2}(s) = \begin{cases} \sigma_{2s}(0) &\text{if } s \in [0,1/2]\\ \sigma_{2(1-s)}(0) &\text{ if } s \in [1/2,1].\\ \end{cases} $$

and finally, for $t \in (1/2,1]$, we have $$ \tau_t(s) = \begin{cases} \sigma_s(0) &\text{if } s \in [0,1-t]\\ \sigma_t\left(\frac{(s+t)-1}{2(2t-1)}\right) &\text{ if } s \in [1-t,t]\\ \sigma_{1-s}(1/2) &\text{ if } s \in [t,1],\\ \end{cases} $$

The only area of concern is at $t=1/2$, and we use the continuity of deformation to handle this. Given $\epsilon > 0$, there exists a $\delta > 0$ such that $|\sigma_t(s) - z| < \epsilon$ whenever $1/2-t < \delta$; consequently, whenever $s \in [1/2 - \delta, 1/2 + \delta]$, it follows that $|\tau_t(s) - z| < \epsilon$.

A similar argument show that there is some $\delta' > 0$ such that $|\tau_t(s)-z| < \delta$ whenever $t - 1/2 < \delta'$.

Hence, by choosing $0 < \eta \leq \min \{\delta, \delta'\}$, it follows that $|\tau_t(s) - z | < \epsilon$ whenever $|t-1/2| < \delta$. Thus, the family $\{\tau_t(s)\}$, as defined above, is a homotopy of paths from $\gamma_1$ to $\gamma_2$, are required.