Yep, you're right. I normally put \\(n\\) before \\(m\\), completely against alphabetical order, so I'm not surprised I slipped up when forced to go the opposite way with \\(F : \mathbf{m} \to \mathbf{n}\\). Thanks for the catch!

Incidentally, we can verify this graphically using the stars and bars method. Notice that stars are input elements, whereas bars separate output elements. So adjacent stars go to the same location in the output space, and each bar increments the "current" output location. Here are all of the functors from \\(\textbf{2}\\) to \\(\textbf{3}\\) in this notation: \*\*||, \*|\*|, \*||\*, |\*\*|, |\*|\*, ||\*\*.