Theorem: For a locally small category $\mathscr{A}$ and $A \in \mathscr{A}$, if $X, Y$ are presheaves with $H_A \cong X + Y$, then either $X$ or $Y$ is the trivial presheaf $\Delta \varnothing$.

Proof: A natural isomorphism $H_A \cong X + Y$ amounts to a universal element $u \in (X+Y)(A) \cong X(A) + Y(A)$. Take an arbitrary $B \in \mathscr{A}$ and $b \in (X+Y)(B) \cong X(B) + Y(B)$. Universality says we have a unique map $\varphi: B \rightarrow A$ with $(X+Y)(\varphi)(u) = b$.

Now either $u \in X(A)$ or $u \in Y(A)$. Suppose that it is in $X(A)$. Then for any map $f : B \rightarrow A$ we have:

$$(X+Y)(f)(u) = (X(f) + Y(f))(u) = X(f)(u)$$

from the behavior of sums of maps in $\mathbf{Set}$. But, this means that $(X+Y)(f)(u)$ is always in $X(B)$. Thus, if we picked a $b \in Y(B)$, there would be no map whose image under $X+Y$ carried $u$ to $b$, violating universality -- thus, there must be no elements in $Y(B)$. Since $B$ was arbitrary, we have $Y(B) = \varnothing$ for all $B$ -- i.e., $Y$ is the constant presheaf $\Delta \varnothing$.

Supposing $u \in Y(A)$ and using a symmetric argument shows that $X = \Delta \varnothing$. Thus, one of $X$ or $Y$ must be the trivial presheaf.