My colleague and I are working through Goldblatt's Topoi, and we're stuck on exercise 2 in chapter 11, section 4.
The exercise is to show that $\|\phi\|^m \circ f = \|\phi\|^m \circ h$, for any formula $\phi$ with $n$ free variables, where $f, h : a^m \to a^n$ are such that $pr^m_{i_k} \circ f = pr^m_{i_k} \circ h = pr^n_k$ for all $1 \leqslant k \leqslant n$.
Here $f$ is defined to be a product arrow whose $j$-th component $p_j$ is defined as follows:
$$ p_j = \begin{cases} pr^n_k : a^n \to a &\text{if } j = i_k \text{, for some } 1 \leqslant k \leqslant n \\ g &\text{otherwise } \end{cases} $$
where $g$ is any arrow from $a^n$ to $a$. Then $h$ is defined to be the same as $f$, except perhaps with a different arrow from $a^n$ to $a$ in place of $g$. The point of the exercise is to show that it doesn't matter whether we define the arrow representing truth in a model $\|\phi\|_{\mathfrak{A}}$ to be $\|\phi\|^m \circ f$ or $\|\phi\|^m \circ h$.
We're on the inductive step where we need to show that $\forall_a \circ |\phi|^m_i \circ f = \forall_a \circ |\phi|^m_i \circ h$, where $|\phi|^m_i$ is the exponential adjoint of $\|\phi\|^m \circ \langle pr^{m + 1}_1, \ldots, pr^{m + 1}_{i - 1}, pr^{m + 1}_{m + 1}, pr^{m + 1}_{i + 1}, \ldots, pr^{m + 1}_m \rangle$.
One strategy we've tried has been to show that $T^{m + 1}_i \circ (f \times \mathsf{1}_a) = T^{m + 1}_i \circ (h \times \mathsf{1}_a)$, so that $|\phi|^m_i \circ f$ and $|\phi|^m_i \circ h$ would be the exponential adjoint of the same arrow, and hence identical, but we haven't been able to show this.