Let $f:X\to \Bbb{R}$ be some continuous function from topological space $X$.Define $\varphi:X'\to X$ be some continuous function also.
If we further assume $\varphi$ is homeomorphism,can we claim $\text{supp}(f\circ \varphi) = \varphi^{-1}(\text{supp}(f)).$
I prove it as follows:Denote $Z_f= \{x\mid f(x) = 0\}$ ,then we can show $f\circ \varphi(\varphi^{-1}(Z_f)) = f(Z_f）,f\circ \varphi(\varphi^{-1}(Z_f^c)) = f(Z_f^c）$
Hence $Z^c_{f\circ \varphi} = \varphi^{-1}(Z_f^c)$ hence taking closure we have the desired result(since $\varphi $ is homeomorphism )
If we do not assume $\varphi$ is homeomorphism ,do we have relation between $\text{supp}(f\circ \varphi) $ and $ (\text{supp}(f)).$ ?
The converse direction is also not so hard,since any $x\notin \varphi^{-1}(\text{supp}(f))$ then exist some neiborhood $U_{\varphi(x)}$ that also not contains in $\text{supp}(f)$ , hence exist neiborhood of $x$ which is $\varphi^{-1}(U_{\varphi(x)})$ does not contains in the support of $f\circ\varphi$.Which means :
$$\text{supp}(f\circ \varphi) \subset \varphi^{-1}(\text{supp}(f))$$
This inclusion is strict in general taking $f(x) = x$ ,$\varphi(x') = 0$
then $\text{supp}(f) = \Bbb{R}$, then we have $\emptyset=\text{supp}(f\circ \varphi) \subset \varphi^{-1}(\text{supp}(f)) = \Bbb{R}$
Is my interpretation correct?
Best Answer
Consider $X=\mathbb{R}$ (with the usual topology) and $f(x)=x$. Then $\operatorname{supp}(f)=\mathbb{R}$.
On the other hand, if $X'$ is the set of real numbers with the discrete topology and $\varphi$ is the identity map, then $\operatorname{supp}(f\circ\varphi)=\mathbb{R}\setminus\{0\}$.