Equivalence of toposes $\mathcal{E}/(X + Y) \simeq \mathcal{E}/X \times \mathcal{E}/Y$

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Notation: I write $A + B$ for the coproduct.

Nota bene: I have no idea if the pseudo-inverses I cooked up are the right ones, but I think those are the only ones you can define.

For $\mathcal{E}$ a topos, I am trying to prove that there is an equivalence $\mathcal{E}/(X + Y) \simeq \mathcal{E}/X \times \mathcal{E}/Y$.

Label the inclusions $X \xrightarrow{i} X + Y \xleftarrow{j} Y$ and define the functors $$\langle i^*, j^* \rangle: \mathcal{E}/(X + Y) \to \mathcal{E}/X \times \mathcal{E}/Y$$ ($i^*$ denotes taking the pullback over $i$) and

the functor $$+: \mathcal{E}/X \times \mathcal{E}/Y \to \mathcal{E}/(X + Y)$$

which sends $(a: A \to X, b: B \to Y)$ to $a + b: A + B \to X + Y$ and morphisms $(f, g): (a, b) \to (a', b')$ to $f + g: A + B \to A' + B'$.

The composition $\langle i^*, j^* \rangle \circ +$ is equal to the identity, I am having trouble proving $\circ + \langle i^*, j^* \rangle$ is isomorphic to the identity functor.