Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $\mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $s\in [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $\bar{s}$ and call $a := f_0(\bar{s})$ and $b := f_1(\bar{s})$. Examining the homotopy, $$ H_{\bar{s}}(t) := F(\bar{s}, t) = a(1-t) + bt $$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $\mathbb{R}^n$. If you draw this line for each $\bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant, $$ \dfrac{dH_{\bar{s}}}{dt} = b - a $$