Push forward of a distribution

Question

Let T be a distribution on a domain $\Omega$ in $\mathbb{R}^n$ and $f:\Omega\mapsto\Omega'$(another domain in $\mathbb{R}^n$) be a smooth map which is also proper on $supp(T)$ (i.e. $f^{-1}(K)\bigcap supp(T)$ is compact for every compact set $K\subset\Omega'$). Then the push-forward $f_{*}(T)$ is a distribution on $\Omega'$ given by $f_{*}(T)(\phi):=T(\phi\circ f)$ for all $\phi\in\mathcal{D}(\Omega')$
In the above, even though $supp(\phi)$ is compact, $supp(\phi\circ f)$ may not be. Then why $T(\phi\circ f)$ is well defined$?$ (I guess $f$ being proper has been assumed to address this issue but I don't see how it is working here).

Answer 1

Let $\chi_\phi \in C_c^\infty(\Omega)$ be identical $1$ on a relative compact neighbourhood of $f^{-1}(supp(\phi)\cap supp(T))$, then $\phi\circ(\chi_\phi\cdot f)$ hast compact support and the value of $T(\phi\circ(\chi_\phi\cdot f)$ does not depend on the choice of $\chi_\phi$.