Question about homeomorphism in continuous function spaces. Let $Tx(t)=(1+t^2)x(1-t^2)$. Is T a homeomorphism for 
a) $T: C[-1,1] \rightarrow C[-1,1]$? 
b) $T: C[0,1] \rightarrow C[-1,1]$?
c) $T: C[0,1] \rightarrow C[0,1]$?
For part a, I tried the following: 
The domain of $x(t)$ is $[-1,1]$, so since the function $1-t^2$ is an even function, in particular, it is symmetric with respect to the y-axis, I claim that the domain of $x(1-t^2)$ is $[0,1]$. Actually, it is $[-1,1]$ but it is symmetric on $[-1,0]$ and $[0,1]$. Therefore, to show this function is not surjective, I tried to show that $t^3$ has no preimage, since it is not symmetric on those intervals. But, I couldn't show it. Am I missing something else or is there any other way of approaching these kinds of questions? 
Thanks for any help.
 A: Let us write $f$ instead of $x$ to avoid confusion.
a) + b) $Tf$ is always an even function. Thus the function $T$ cannot be surjective because there are non-even functions in $C([-1,1])$.
c) You do not say anything about the topology on $C = C([0,1])$, so let us assume that $C$ is considered as a normed linear space (with the $\sup$-norm).


*

*Let $u \in C$. Then the map
$$\phi_u : C \to C, \phi_u(f) = u\cdot f$$
is linear (easy to verify!) and continuous. The latter is well-known, but let us prove it for the sake of completeness. In fact with $c = \lVert u \rVert$ we get
$$\lVert \phi_u(f) \rVert = \sup_{t \in [0,1]} \lvert (u \cdot f)(t) \rvert = \sup_{t \in [0,1]} \lvert u(t) \rvert \cdot \rvert f(t) \rvert \le \lVert u \rVert \cdot \lVert f \rVert = c \lVert f \rVert.$$

*If $u$ does not have a zero, then $\phi_u$ is a homeomorphism. In fact, the function $v = 1/u$ is in $C$ and we have $\phi_v \circ \phi_u = id$ and $\phi_u \circ \phi_v = id$.

*Let $w : [0,1] \to [0,1]$ be continuous. Then the map
$$\psi_w : C \to C, \psi_w(f) = f \circ w$$
is linear (again easy to verify!) and continuous. To see the latter note
$$\lVert \psi_w(f) \rVert = \sup_{t \in [0,1]} \lvert (f(w(t)) \rvert \le \sup_{s \in [0,1]} \lvert f(s) \rvert = \lVert f \rVert.$$

*Let $w : [0,1] \to [0,1]$ be a homeomorphism. Then $\psi_w$ is a homeomorphism. In fact, if $w^{-1}$ is the inverse homeomorphism, then $\psi_{w^{-1}} \circ \psi_w = id$ and $\psi_w \circ \psi_{w^{-1}} = id$.

*Let $u(t) = 1+t^2$ and $w(t) = 1-t^2$. The latter is a homeomorphism with inverse $w^{-1}(t) = \sqrt{1-t}$. We have
$$T = \phi_u \circ \psi_w$$
which proves that $T$ is a homeomorphism.
