Suppose $X$ is a space and $\pi_1(X, x)=\{e\}$, the trivial group. Show that there is a homotopy

Question

The condition is that $\gamma_0,\gamma_1$ are paths in $X$ such that $\gamma_0(0)=\gamma_1(0)=x$ and $\gamma_0(1)=\gamma_1(1)=y$, then there is a homotopy $\{f_t\}_{t\in I}$ with $f_0=\gamma_0,f_1=\gamma_1$ and $f_t(0)=x, f_t(1)=y, \forall t\in I$.

My instructor says it is quite obvious but I don't even know what to begin doing. We just started learning about homotopy and I am struggling with this exercise, would appreciate some detailed explanations.

Answer 1

Let $F: [0,1] \times [0,1] \to X$ be a homotopy so that $F(0,0) = x$, $F(1, 1) = y$ and the left and top edge corresponds to $\gamma_1$, while the right and bottom edge corresponds to $\gamma_2$. That is, $$\begin{array}{cc} F(0, t) = \gamma_1(t/2), & F(t, 1) = \gamma_1(1/2 + t/2), \\ F(t, 0) = \gamma_2(t/2), & F(1, t) = \gamma_2(1/2 + t/2). \end{array} $$

Such a homotopy exist because the four edge of $[0,1] \times [0,1]$ forms a loop, and such a loop is homotopic to the trivial map as $X$ is simply connected. Now define $t\in I = [0,1]$ and

$$f_t (s) = \begin{cases} F\big((1-2s)(0,0) + 2s(t, 1-t)\big) & \text{if } s\in [0,1/2],\\F\big( (2s-1)(1,1) + (2-2s)(t, 1-t) \big) & \text{if } s\in [1/2,1].\end{cases} $$

(The first half is a line connecting $(0,0)$ and $(t,1-t)$, while the second half connects $(t,1-t)$ to $(1,1)$)

So $f_t(0) = x$, $f_t(1) = y$ for all $t$ and $\{f_t\}$ is a homotopy connecting $\gamma_1$ and $\gamma_2$.

Answer 2

You're given paths $f,g:I\to X$ with source $x$ and target $y$, so the path $f\ast g^{-1}$ has source and target $x$, i.e. it is a loop based at $x\in X$. You know that $\pi(X,x)$ is trivial, so this loop is nullhomotopic: $[f\ast g^{-1}]=[\varepsilon_x]$, and the homotopy is relative to $\partial I$. It suffices you show that this implies $f$ and $g$ are homotopic relative to $\partial I$.

More generally: suppose $f,g:I\to X$ are loops based at $x$ that are homotopic and suppose $h:I\to X$ is a path from $x$ to $y$. Then $f\ast h\simeq g\ast h$ relative to $\partial I$. Indeed, if $f_t : I\times I\to X$ is a homotopy relative to $\partial I$ from $f$ to $g$, consider $f_t'= f_t\ast h$ (with the obvious meaning). You need to show this is continuous, and this gives the desired homotopy.

To see how this finishes the proof in the first paragraph, $f\ast g^{-1}\simeq \varepsilon_x$ gives $$f\simeq f\ast g^{-1}\ast g\simeq \varepsilon_x \ast g\simeq g$$ and you're done, since we know these homotopies can be taken relative to $\partial I$.