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Suppose there are variable $x, y, t$ where $x=f(t)$ and $y=g(t)$ where $f$ and $g$ are functions* of $t$.

Is there any circumstance under which it is necessarily true that $y$ can be represented as a function* of $x$?

*: the type of function that $f$ and $g$ are presumably affects whether or not the proposition is true, and possibly the type of function relating $x$ and $y$ as mentioned in the proposition itself.

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    At minimum, you need that if $f(t_1)=f(t_2)$ then $g(t_1)=g(t_2)$. – 2017-02-16
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    @ThomasAndrews: could you please elaborate and explain the reasoning of this in an answer? – 2017-02-16
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    It wasn't an answer, so I can't write an answer with it. It is a requirement for $h$ to exist, but, depending on the conditions on $f,g,h$ - say, continuity or differentiability - it is not enough to ensure that $h$ exists, so this condition does not match what you asked for. – 2017-02-16

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Yes, there are conditions. Mainly, $f(t)$ must be one-to-one if $g(t)$ is one-to-one. That is,

$$g(a)\ne g(b)\implies f(a)\ne f(b)$$

Take $f(t)=(t-2)^2$ for example. It is then true that $f(1)=1=f(3)$ when $a=1$ and $b=3$. This means that $y=g(1)=g(3)$, but as we said above, $g(1)\ne g(3)$. Hence, $y$ cannot be written as a function of $x$ here.

(To restate the above, when $x=1$, it is both possible for $t=1$ and $t=3$, but these values of $t$ will give different values of $y$, so there is no single $y$ such that $y=h(x)$.)

However, if $f(a)=f(b)\implies g(a)=g(b)$, then $y$ can be written as a function of $x$.