How to show that the curves $\gamma_0 , \gamma_1$ are homotopic with conservation of endpoints? Given are the curves $\gamma_0 , \gamma_1: [0, 1] \longrightarrow \mathbb{R}^2$ which are defined as:
$\gamma_0(t) = (t,0)$ and $\gamma_1(t) = (t,f(t))$
with continuous function $f: [0, 1] \longrightarrow \mathbb{R}^2$ defined as $f(0) = f(1) = 0$.
Next, I need to show that the curves are homotopic with endpoints conserved. I know that for a homotopy between $\gamma_0 , \gamma_1$ we mean that we have a continuous map $\Gamma: [0,1] \times I \longrightarrow \mathbb{R^2}$ such that $\forall s \in [0,1]: \space \Gamma (s, 0) = \gamma_0(s)$ and $\space \Gamma (s, 1) = \gamma_1(s)$. But besides a homotopy, the 2 curves have common end points, ie: $\gamma_0(0) = \gamma_1(0)$ and $\gamma_0(1) = \gamma_1(1)$.
We say that the homotopy retains start and end points if for all $t \in [0,1]$ (with $I=[0,1]$)$\space$:
$\Gamma(0,t) = \gamma_0(0)$ and $\space \Gamma(1,t) = \gamma_0(1)$ and of course this also applies to $\gamma_1$ instead of $\gamma_0$.
From here I get stuck. In my view (and I see this in other examples on the internet too), they define $\Gamma(s,t)$ to show that there is a homotopy preserving endpoints of the curves. However, I don't quite understand how you should just set up that $\Gamma(s,t)$ yourself.
Thanks in advance for your time and help! And maybe you could give me some tips/hints so I will be able to set these up myself.
 A: Drawing the problem helps. Essentially you have the graph of a continuous function $f$ on $[0,1]$ that vanishes on the boundary, and you want to gradually expand the graph of the zero function up to this graph. The homotopy $$ \Gamma(s,t) = (s,tf(s)) $$ surely does the job, as when $t$ gradually goes from $0$ to $1$ you gradually go from the curve $\gamma_0(s) = \Gamma(s,0) = (s,0)$ to the curve $\gamma_1(s) = \Gamma(s,1) = (s,f(s))$.
A: Your goal here, when constructing $\Gamma (s,t)$ is to create a continuous function where, when $s=0$, $\Gamma$ returns $\gamma_0$ and when $s=1$, $\Gamma$ returns $\gamma_1$. Moreover, $\Gamma$ itself needs to be a continuous function, so that $s$ effectively parameterizes the homotopy between $\gamma_0$ and $\gamma_1$. Note that for any fixed $t_0$, the map
$$\Gamma_{t_0}(s,t_0)=(1-s)\cdot 0+s\cdot f(t_0)=sf(t_0)$$
is a continuous function (in fact, it's just a line segment). I leave it the reader to show that $\Gamma$ itself is continuous. In general, you typically can let your homotopy look something like
$$\Gamma (s,t)=(1-s)\gamma_0+s\gamma_1$$
and it should work out. The idea here is that your continuously transforming each point along $\gamma_0$ to it's corresponding point on $\gamma_1$ (this transformation is, in fact, linear for any fixed $t_0$ and so surely continuous for $t_0$).
