Let $I=[0, 1]$. I need some help to define a continuous function $f:I\times I\longrightarrow I$ such that $$f(0, t)=t\quad \textrm{and}\quad f(1, t)=1$$ for all $t\in I$. The nearest I got was: $$f(s, t)=\frac{t}{(t-1)s+1},$$ but this function has a singularity when $s=1$ and $t=0$.
Motivation: The motivation for this question comes from algebraic topology:
Let $X$ be a topological space and $I=[0, 1]$. Consider the cylinder $X\times I$ and collapse $X\times \{1\}$ into a point. The space obtained is called cone, denoted by $C(X)$. If I find $f:I\times I\longrightarrow I$ continuous with those properties I'll be able to show $C(X)$ is contractible.
$$ f(s,t) = (1-s)t + s $$
Now, it might seem a little bit random coming up with this, but I basically just treated it like an interpolation problem where $s$ is the variable and we have two data points $(0,t)$ and $(1,1)$.
I'm not sure if there is a context for this question but, to me, this looks pretty good.
So if we had a more general problem
$$ f(a,t) = g_1(t),f(b,t) = g_2 (t) $$
then basically my thought process was saying okay, when $s=0$ and $s=1$ we want this function to behave differently, so lets consider a general function
$$ f(s,t) = (s-1){\color{Red} {g_1}} + (s-0){\color{Red} {g_2}} \hspace{25 mm} (*) $$
where these function $g_1$ and $g_2$ are what we want to find.
So we know when $s=0, f(s,t) = t$, so we get, by substituting this into $(*)$ $t = -g_1$, and we do the same thing when $s = 1$ to get $g_2 = 1$.
I hope this explains my thought process and the method involved here, its quite a simple trick, but you might like to try and do the same thing to find a continuous function $f(x)$ that satisfies:
$$ f(n) = 2^n \text{ for n} = 1,2,3,4,5 \text { and } f(6) = \pi $$
And post your answer in the comments or something, its a bit more difficult but hopefully it'll be okay.