Node voltage and actual current path when output is grounded ?

[MathWizard]

When I've been finding things like Y or T parameters, I've been doing node analysis on the shorted output node, and getting the right answer until yesterday (at least for simple cases).  And I see it done by prof's in pdf's too, like on page 3.
https://d13mk4zmvuctmz.cloudfront.net/assets/main/study-material/notes/electrical-engineering_engineering_network-analysis_two-port-network_notes.pdf

Normally you would setup up at least 1 less equation than total node #, and not setup the equation for the reference node. When I setup the shorted output node eqn, I don't include everything else that actually enters/leaves the reference node.

So if you make a circuit like below, similar to a small signal model BJT with a feedback path between base-collector, and short the output, I would say the current from R2 still enters physical node c, and the correct amount goes into B1, and the rest takes the output V2 path. And if I replace the perfect wire with any amount of small resistance, you get a slightly smaller output current.

So in this pdf below about BJT's and their capacitance, calling i_out the actual current that goes to the collector pin.....won't their simplified circuit give the wrong i_out ? The actual, dependent current source current, will still be the same. But how can say the current through Cu, would not enter the collector region, and then partly out the collector to AC ground ??
see page 14 or pic below
https://pages.hmc.edu/mspencer/e151/sp21/notes/16.pdf

[MathWizard]

I've gone through a bunch of 2-port problems with dependent sources, in a pdf for review, and got them all right.

So for the BJT , in the top of their equation, how can they not be missing a -sCu term ??

i_out= i_b[(gm-sCu)/(gpi+s(Cpi+Cu))]

[MathWizard]

So I am right, it's just they ignore it for the frequency and typical gm the model is meant to work at.