Why junction removal method doesn't work like this? This is a question from my textbook's chapter on electrical currents. It has a square circuit as shown in figure 1, where all the wires have identical resistivity. 
While solving the circuit the author removed the junction as in figure 2, and said it's all due to symmetry.
I have two questions:  


*

*What the symmetry has to do with the current.  

*If there is really no current in the junction, why can't it break like I have drawn in figure 3?



 A: 
What the symmetry has to do with the current

The symmetry used in the operation is not applied to the current but to the arrangement of the resistors in your particular circuit. The current starting from point 2 will travel along two different direction as 2$\rightarrow$C(centre) and 2$\rightarrow$4. At the centre two currents enter, one from the point 4 and another from the point 2. This circuit has a symmetry about the points 1 and 2, such that whatever the current that passes into the 2$\rightarrow$C will be the same current to move from C$\rightarrow$1 and the same holds for the current from 4 and the current towards 3. This is a consequence of the symmetry about the points 1 and 2 which makes it essential that the current distribution be unaltered even if you rotate the circuit about the perpendicular bisector of line$1\rightarrow 2$. Therefore, at the center the net effect is as if the connection is not valid and can be broken as in figure 2 since there is no "exchange" of current between the two parts into which the junction is split in figure 2, the current from 2 to the center moves independently of the current from 4 towards 1.   

If there is really no current in the junction, why it can't break like
  I have drawn   

There is current in the junction but it is just the division which makes removal of the connection for calculation sake plausible. If you divide it the way you do in figure 3, you are not exploiting the symmetry and hence the divided circuit is not the same as the original circuit with the intact junction. The current of the 2$\rightarrow$C and the C$\rightarrow$1 branches is the same and not 2$\rightarrow$c and C$\rightarrow$4. If the later had been true, figure 3 would have been correct but in this case figure 2 is correct.
