I am trying to prove a theorem of Quackenbush (Theorem 3.8 in Chapter V of Burris & Sankappanavar):
If $V$ is a locally finite variety with only finitely many finite subdirectly irreducible members (up to isomorphism), then $V$ has no infinite subdirectly irreducible members.
The proof ends with the following situation: Let $A$ be any algebra in $V$, and $A \in ISP(V^*)$, where $V^*$ are the finitely many finite s.i. algebras. From this it should follow that $A \in IP_S(V^*)$. This is stated without further comment, so I suppose I am overlooking something very easy. How do I prove this fact?
We have $A \subseteq \prod_{i\in I} B_i$, with $B_i\in V^*$ for all $i\in I$. For all $i$, let $C_i = \pi_i[A] \subseteq B_i$, where $\pi_i$ is the projection to $B_i$ (so $\pi_i$ is surjective onto $C_i$). Let $C_i \subseteq \prod_{j\in J_i} D_{i,j}$ be the subdirect representation of $C_i$ by its subdirectly irreducible quotients, and again let $\pi_j$ be the projection to $D_{i,j}$ (this time we know the $\pi_j$ are surjective onto $D_{i,j}$). Since each $B_i$ is finite, each $C_i$ is finite, so all the $D_j$ are finite, and hence they're all in $V^*$.
Now for all $i\in I$ and $j\in J_i$, the map $\pi_j\circ \pi_i\colon A\to D_{i,j}$ is surjective. And for $a\neq b$ in $A$, there is some $i\in I$ such that $\pi_i(a) \neq \pi_i(b)$ in $C_i$, and there is some $j\in J_i$ such that $\pi_j(\pi_i(a)) \neq \pi_j(\pi_i(b))$ in $D_{i,j}$, so the induced map from $A$ to $\prod_{i\in I}\prod_{j\in J_i} D_{i,j}$ is an embedding.