How would this transistor diagram be drawn if we draw it with all the closed circuits?
Your first picture is almost correct, but you can't connect the top and ground parts directly. If you would connect them you would build a short circuit and you had no voltage over your transistor circuit anymore.
The way the diagram in the book is drawn is a typical way to show where the (+) and the (-) of the voltage source is connected. I have drawn a battery (red) in the circuit to show what exactly that means. At the top you have the positive voltage of say 1.2V and the ground symbols are connected to the negative terminal and are interpreted as 0V.

C is not connected to 1.2V or ground at the output (like you have drawn it), but by the transistors. That's the whole purpose of this circuit. C is the output that (dependent on the input signals) is either connected by the upper two transistors to the positive terminal (1.2V) or the negative terminal (0V). I have drawn these two paths in blue and green.
This way the circuit fulfills the role as a logic gate: The two inputs A and B both have to be at LOW potential for the upper two transistors to be conductive. So with both inputs LOW, the output C is connected to logic HIGH (the supply voltage). The lower two transistors are not conductive in this case. But if one of the inputs is at HIGH potential, one of the two upper transistors does not conduct anymore, so the output is not connected to the supply. Instead one of the lower transistors gets conducting and the outputs is &quot;pulled LOW&quot;.
The circuit is representing a NOR gate.