Let $f$, $g$ be functions from $[0,1]$ to $[0,1]$ with $f$ strictly increasing . Then
$A.$ If $f$ is continuous then so is $f\circ g.$
$B.$ If $f$ and $f\circ g$ are continuous then so is $g.$
$C.$ If $g$ and $f\circ g$ are continuous then so is $f.$
Now for $A$ , I can take $g(x)=x$ and $f$ any discontinuous function thus this is not the case.
If $g$ is the Dirichlet function and $f$ is the $0$ function or $f(x)=1$ for all $x\in [0,1]$ then $B$ could be cancelled but $f$ needed to be strictly increasing here so this was nonsense .
So I am not sure about the last two options.
Any help is appreciated . Thanks.
B is true. Note that there exists a continuous function $h$ such that $h(f(x))=x$ for all $x\in [0,1]$.
C is false. Take $g$ to be a constant function, and construct a suitable counterexample from there.
Proof that B is true: Suppose that $f$ and $g$ satisfy the hypothesis of B. Because $f$ is continuous and strictly increasing, there exists a continuous function $h:[0,1]\to [0,1]$ such that $h(f(x)) = x$ for all $x \in [0,1]$.
Now, since $f \circ g$ is continuous, it must be that $h \circ (f \circ g)$ is continuous. However, note that for all $x$ we have $h(f(g(x))) = g(x)$, so that $h \circ f \circ g = g$. Thus, $g$ must be continous.
Proof that C is false: Take $f(x)$ to be the function $$ f(x) = \begin{cases} x/3 & x \in [0,1/2)\\ 2x/3 & x \in [1/2,1] \end{cases} $$ take $g(x) = 1$. This $f$ and $g$ provide a counterexample to the claim given.