I was hoping to know the difference between equation (1) and equation (2). Would they be considered equal or is equation (2) less strict compared with equation (1)?
Let the correspondence $F:\mathcal A→2^\mathcal A$ be defined as follows:
$F(\mathbf a)=f_1(\mathbf a_{-1})\times...\times f_ \mathcal K(\mathbf a_{-\mathcal K})$.
$\exists\mathbf a\in A: \mathbf a\in F(\mathbf a)$ (1)
$\forall k \in \mathcal K, \exists \mathbf a \in \mathcal A: a_k \in f_k(\mathbf a_{-k})$ (2)
Thank you.
The second statement is less restrictive.
The first statement says that there exists some $\mathbf a\in A$ such that $a_1\in f_1(\mathbf a_{-1})$ $\ldots$ $a_\mathcal K\in f_{\mathcal K}(\mathbf a_{-\mathcal{K}})$.
The second statement says that for each $k\in\mathcal K$ there exists some $\mathbf a\in A$ (and it is possible that this $\mathbf a$ is different for each $k\in\mathcal K$) such that $a_k\in f_k(\mathbf a_{-k})$.
It may be more illuminating to write (2) as follows: $$\forall k \in \mathcal K, \exists \mathbf a^k \in \mathcal A: a_k^k \in f_k(\mathbf a_{-k}^k)$$ to emphasize with a superscript that $\mathbf a$ need not be common across $(1,\ldots,\mathcal K)$, unlike in (1).