Does there exist a continuous $g(x,t)$ such that every continuous$ f(x)$ equals $g(x,t)$ for some $t$ and all $x$?? Is there a continuous $g(x,t)$ such that every continuous $f(x)$ equals $g(x,t)$ for some $t$ and all $x$?
$f$ is from $[0,1]$ to itself with $f(0)=0$ and $f(1)=1$ and is either smooth or continuous (your choice)
$g$ is defined on $[0,1]\times(0,1)$ or $[0,1]\times[0,1]$ (your choice)
What if we require there to be a unique $t$ for each $f$?
Is there an simple explicit such $g$?
 A: Update: I've modified my answer to account for the boundary conditions $f(0)=0$ and $f(1)=1$. Such continuous functions  restricted to $I_2$ give all of $C(I_2)$.
For $n>1$, define $I_n=[1/2^n,1-1/2^n]$. Since $g$ is uniformly continuous on $I_n\times I_2$, the map $\Phi_n: I_n\to C(I_2)$ by $t\mapsto g(t,x)|_{I_2}$ is continuous from $I_n$ into the Banach space $C(I_2)$.
The set of functions $(g(t,x)|_{I_2})_{0<t<1}=\cup_n \Phi_n(I_n)$ is $\sigma$-compact in $C(I_2)$ and therefore a strict subset of $C(I_2)$. Thus $(g(t,x))_{0<t<1}$ is a strict subset of $\{f\in C[0,1]: f(0)=0, f(1)=1\}.$
A: Let $X$ denote the space of continuous maps $f:[0,1]\to[0,1]$ such that $f(0)=0$ and $f(1)=1$, with the uniform metric.  Then $X$ is a complete metric space, and it is easy to see that every compact subset of $X$ has empty interior.  In particular, it follows by the Baire category theorem that $X$ is not $\sigma$-compact.
Now let $g:[0,1]\times(0,1)\to[0,1]$ be any continuous map and let $\hat{g}:(0,1)\to X$ denote the induced map $t\mapsto (x\mapsto g(x,t))$.  Then $\hat{g}$ is continuous (this follows from a simple argument using compactness of $[0,1]$).  Since $(0,1)$ is $\sigma$-compact, the image of $\hat{g}$ is also $\sigma$-compact.  In particular, $\hat{g}$ cannot be surjective.
This settles your question when $f$ is required only to be continuous.  In the smooth case,  the same argument applies, because by a minor modification of the argument in this answer, the subset of $X$ consisting of all smooth maps is not $\sigma$-compact either.
A: If $f$ is allowed to be only a polynomial, than the answer is yes, but the $g$ is a really wild function.

Start with some notation. Let $\mathcal{P}_n$ be the space of polynomials of degree up most $n$. We will use a lot the fact that $\mathcal{P}_n$ is isomorphic to $\mathbb{R}^{n+1}$.
Let $s_n:[0,1] \rightarrow [-1,1]^n \subset \mathbb{R}^n \cong \mathcal{P}_{n-1}$ be a space-filling curve such that $s_n(0)=s_n(1)=0$.
With all set, we can start constructing function $g$. First we construct function $G:[0,\infty) \rightarrow \mathbb{R}^{\mathbb{N}}$.
$$
G(t) = n s_n(x-n) \qquad \text{for }n\leq x < n+1
$$
The last thing to do is to reinterpret $G$. Image of $G$ is the space $c_{00}$, the space of sequences with finitely many non-zero elements, which is isomorphic to the space of all polynomials $\mathcal{P}$, thus we can understand $G$ as function $[0,\infty) \rightarrow \mathcal{P}$. Let $h:(0,1)\rightarrow [0,\infty)$ be continuous surjective function. Then we define $g$ as
$$
g(x,t) = G(h(t))(x)
$$

This can be generalized. If $f\in X$, where $X$ is function space which can be written out as countable union of finitely dimensional subspaces, then the answer is yes. We can proceed in the same way and construct continuous curve $\gamma: (0,1)\rightarrow X$ which fills the whole space. 

I guess, than the question can be answered fully if you find out what is know about space-filling curves, unfortunately I do not know much about them and don't have the time to look it up right now.
