I’ve added the following to my research book:

Definition
Galois surjection is the special case of Galois connection such that $f^{\ast} \circ f_{\ast}$ is identity.

Proposition
For Galois surjection $\mathfrak{A} \rightarrow \mathfrak{B}$ such that $\mathfrak{A}$ is a join-semilattice we have (for every $y \in \mathfrak{B}$)

$f_{\ast} y = \max \{ x \in \mathfrak{A} \mid f^{\ast} x = y \}.$

(Don’t confuse this my little theorem with the well-known theorem with similar formula formula $f_{\ast} y = \max \{ x \in \mathfrak{A} \mid f^{\ast} x \leq y \}$.)

This formula in particular applies to the Galois connection between funcoids and reloids (see my book).

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I have defined two new kinds of products of funcoids:

1. $\prod^{\mathrm{in}}_{i \in \mathrm{dom}\, f} f = \prod^{(C)}_{i \in \mathrm{dom}\, f} (\mathsf{RLD})_{\mathrm{in}} f_i$ (cross-inner product).
2. $\prod^{\mathrm{out}}_{i \in \mathrm{dom}\, f} f = \prod^{(C)}_{i \in \mathrm{dom}\, f} (\mathsf{RLD})_{\mathrm{out}} f_i$ (cross-outer product).

These products are notable that their values are also funcoids (not just pointfree funcoids).

See new version of my book for details.

I’ve added to my book the following conjecture:

Conjecture For every composable funcoids $f$ and $g$

$(\mathsf{RLD})_{\mathrm{out}}(g\circ f)\sqsupseteq (\mathsf{RLD})_{\mathrm{out}}g\circ(\mathsf{RLD})_{\mathrm{out}} f.$

After noticing an error in my math book, I rewritten its section “Funcoids and filters” to reflect that $(\mathsf{RLD})_\Gamma = (\mathsf{RLD})_{\mathrm{in}}$.

Previously I proved an example demonstrating that $(\mathsf{RLD})_\Gamma \ne (\mathsf{RLD})_{\mathrm{in}}$, but this example is believed by me to be wrong. The example was removed from the book.

Thus I removed all references to $(\mathsf{RLD})_\Gamma$ (as it is the same as $(\mathsf{RLD})_{\mathrm{in}}$) and reworked the chapter “Funcoids and filters” to reflect the change.

The book is available free of change at this Web page.

The story of the past:

$(\mathsf{RLD})_\Gamma$ was defined by the formula $(\mathsf{RLD})_\Gamma f = \bigsqcap^{\mathsf{RLD}} \mathrm{up}^\Gamma\, f$.

From the theorem in “The diagram” section (the theorem with a diagram) it trivially follows that $(\mathsf{RLD})_\Gamma f = (\mathsf{RLD})_{\mathrm{in}} f$. It follows trivially, but I have found this only today.

I proved both $(\mathsf{RLD})_\Gamma \ne (\mathsf{RLD})_{\mathrm{in}}$ and $(\mathsf{RLD})_\Gamma = (\mathsf{RLD})_{\mathrm{in}}$.

So there is an error in my math research book.

I will post the details of the resolution as soon as I will locate and correct the error. While the error is not yet corrected I have added a red font note in my book.

I claimed earlier that I partially solved this open problem.

Today I solved it completely. The proof is available in this PDF file.