Verify that if $f_* : \mathbb{R}^J \rightarrow \mathbb{R}^I$ and $g_* : \mathbb{R}^I \rightarrow \mathbb{R}^J$ are defined by $$f_*(g)_i = -\epsilon \log\left(\sum_{j \in J} e^{(g_j - C_{ij})/\epsilon} \beta_j\right) \text{ and } g_*(f)_j = -\epsilon \log\left(\sum_{i \in I} e^{(f_i - C_{ij})/\epsilon} \alpha_i\right)$$ then for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, we have $\frac{\partial G}{\partial f_i}(f_*(g), g) = \frac{\partial G}{\partial g_j}(f, g_*(f)) = 0$ for all $(i,j) \in I \times J$.
Verify that if $f_* : \mathbb{R}^J \rightarrow \mathbb{R}^I$ and $g_* : \mathbb{R}^I \rightarrow \mathbb{R}^J$ are defined by
$$f_*(g)_i = -\epsilon \log\left(\sum_{j \in J} e^{(g_j - C_{ij})/\epsilon} \beta_j\right) \text{ and } g_*(f)_j = -\epsilon \log\left(\sum_{i \in I} e^{(f_i - C_{ij})/\epsilon} \alpha_i\right)$$
then for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, we have $\frac{\partial G}{\partial f_i}(f_*(g), g) = \frac{\partial G}{\partial g_j}(f, g_*(f)) = 0$ for all $(i,j) \in I \times J$.