There seems to be a temptation to underestimate simple and small examples, let me therefore explain.
Letís draw the tree belonging to this 4◊3 board position. The page count nicely explains that. Black goes down to the left, Right to the white, oops. Counts (from Blackís perspective) are given at the nodes.
0 0 / \ / \ / \ / \ / \ b / \ a a / \ b / \ / \ 5 -5 5 -5 a / \ / \ b b / \ / \ a / \ / \ / \ / \ 10 0 0 -10 10 0 0 -10 c/ \d c/ \d 1 -1 1 -1
Actually this is one tree, roots joined. The counts above the leafs were made by averaging. This only was a first guess, but luckily the temperature (distance between mother and daughter) going down never rises, sparing us a recalculation.
Donít confuse the situation above with the 4◊5 board one below.
Its count is 0 too
(due to symmetry, of course),
but haneís miai value only is 1,
setting the temperature, and
descentís miai value only is Ĺ.
I donít know about the temptation you mentioned, but I know that I fell for the temptation of assuming that you would not do
even though thereís barely any difference in complexity for your small example.
(I was assuming that your diagram gave a region in which the already-present stones are immortal and the left will be white territory to the extent that Black doesn't penetrate underneath from the right and similarly for the right being black territory.
For such a region, the comment I left with my revert would've been correct.)
Also, since your calculation came up with a non-zero temperature (namely, 5) such that
under alternating play, neither player can get more moves at that temperature,
what your calculation found is the temperature at which the mast changes from purple (below) to something else (above).
In your example, I agree that play here should be at temperature 5, since whoever begins here chooses what the resulting infinitesimal will be if the opponent gets the next play here, so I agree that your example is an Ambiguous Position and that in it, the descent and the hane are both ambiguous.
However, for other (loopless) positions, I believe your method can get a temperature which is higher than the temperature at which the position should be played.
Let UP be
/\ / \ 1 / \ / \ 0 -2
and let DOWN be the negative of that.
The result of joining the roots of
/\ /\ /\ / \ / \ / \ / \ 2 -2 / \ / \ / \ / \ / \ / \ / \ /\ / \ /\ / \ / \ / \ / \ / \ 10 DOWN UP -10 8 UP DOWN -8
should be played at temperature 4, even though I believe your method would conclude its temperature is 5.
 That's why I once suggested to have round corners. They are! Zoom in. It's a full-board example.
 Great that we agree!