The map $\phi^{\star}$ is injective since for any two characters $\chi_{1},\chi_{2}:\mathcal{A}_{2}\to K$, if $\chi_{1}\circ\phi=\chi_{2}\circ\phi$, then the restriction of $\chi_{1}$ and $\chi_{2}$ on the image of $\phi$ are equal, hence the two characters are equal by the density of image.

Let $(z,\lvert\mathord{\cdot}\rvert_{z})\in\mathfrak{M}(\mathcal{A}_{1})$ which is not in the image of $\phi^{\star}$, then $\ker(\phi)\nsubseteq\mathfrak{p}_{z}$: otherwise the character $\mathcal{A}_{1}/\ker(\phi)\to\hat{\kappa}(z)$ extends to a character $\mathcal{A}_{2}\to\hat{\kappa}(z)$ by the density of image of $\phi$. Now there exists $f\in\ker(\phi)\setminus\mathfrak{p}_{z}$, so $|f|_{z}\neq 0$. For small enough $\epsilon>0$, the basic open set $U(f;|f|_{z}-\epsilon,|f|_{z}+\epsilon)\subset\mathfrak{M}(\mathcal{A}_{1})$ is a neighbourhood of $(z,\lvert\mathord{\cdot}\rvert_{z})$ which is not contained in the image of $\phi^{\star}$. So the image of $\phi^{\star}$ is a closed subset in $\mathfrak{M}(\mathcal{A}_{1})$. ∎