= Solution
Take two <subobjects> $A,B\hookrightarrow X$ whose inverse images coincide. Their characteristic maps $a,b:X\to\Omega_{\mathcal E}$ satisfy $\bar\lambda f^*a=\bar\lambda f^*b$ by part (i). Under the <adjunction> $f^*\dashv f_*$, these maps transpose to $\lambda a$ and $\lambda b$, respectively, by naturality of the unit. If $\lambda$ is a <monomorphism>, then $a=b$ and hence $A=B$. Thus $f^*$ reflects equality of subobjects.
For parallel arrows $r,s:X\rightrightarrows Y$ with $f^*r=f^*s$, their <equalizer> $K\hookrightarrow X$ pulls back to all of $f^*X$. Reflection of equality with the whole subobject gives $K=X$, so $r=s$. Therefore \b[the inverse image is faithful, and the geometric morphism is surjective]. This is the <surjectivity detected by the subobject classifier> criterion. For completeness, faithfulness is equivalent here to conservativity: if $f^*h$ is invertible, the image inclusion of $h$ and its kernel-pair diagonal become whole subobjects; reflection makes $h$ both a regular epimorphism and monic, hence invertible in the <topos>.
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