How would one go about constructing a covariant power object functor (like powerset functor in category Set).
Obviously, on objects, the mapping would equal $$ A \mapsto \Omega^A. $$
But what about morphisms? I've looked about the definition of power object (https://ncatlab.org/nlab/show/power+object), and the only sensible idea that came to mind is taking the diagram from the definition (I'm having trouble drawing commutative diagrams on here), replacing $r \hookrightarrow c \times d$ with $m \times \text{id}_{\Omega^d}:\text{im} f \times \Omega^d \hookrightarrow c \times \Omega^d$, where $f:d \to c$ is the morphism I want to look at, and $m$ is its image $m:\text{im} f \hookrightarrow c$.
That way i get the unique morphism $\chi_r:\Omega^d \to \Omega^c$. I'm not quite sure if that is the case, but it seems like if i get a pair $x \times X$ of the type $d \times \Omega^d$ and map it via morphism $g \times \text{id}:d \times \Omega^d \to c \times \Omega^d$, then $g(x) \in_c \chi_r(X)$ only if $g$ factors through image of $f$.
However, I fail to see the functor property of it. When trying to prove it, I only seem to be able to come to a morphism $v:\text{im}gf \to \text{im}g$, via the image property and it gets me nowhere close to $\chi_g \chi_f = \chi_{gf}$.
I wasn't able to use pullback properties, since, even though I can construct some morphisms to $\in_c$, I cannot verify the commutativity needed to produce the morphism to the pullback.
I've also looked for answers online extensively and found no positive or negative results.
Is my intuition on using the image wrong (and of so, which monomorphism should I use instead)? Or did I just miss something that can help me to prove it?