Is this what you mean when relating the parity of centre pieces and edge pieces?
The parts about the centers are not correct for the 2nd and 3rd pictures, and only partially correct for the 1st picture.
You have to consider what kinds of centers are cycled, meaning from which orbits. Remember that from a theory standpoint there are two kinds of centers on a supercube (and yes Maarten is right that any conclusion we're drawing here is really only valid on a supercube). When you turn an inner slice on a 4x4x4 cube you have done two 4-cycles of pieces, but you have to consider which center orbit those 4 cycles occur in.
On the 5x5x5 picture you are correct that you have done three center 4-cycles, but without determining how many cycles happen in each kind of center orbit, we can't make any conclusion about how parity of those orbits are affected. Same thing for the 6x6x6 picture.
So once the centres are solved on an odd-dimensional cube, we can guarantee that the edges must be in an even parity state, because the centres must be.
If we're considering regular cubes (and not supercubes), then no - and I'll give you a counterexample as to why not. Say your scrambling algorithm on a 5x5x5 uses an odd number of inner slice quarter turns, and you completely solve the centers using an even number of inner slice quarter turns. The centers are now solved, but what is the parity of the wing edge orbit?
If you're considering supercubes, and you solve all centers to an even parity state, then yes all wing edge orbits must be in an even permutation.
If the centres are in an odd parity state then they cannot be solved.
Sure they can. Keep in mind you are talking supercube centers right now. The way to toggle parity in a supercube "oblique" center orbit is by either turning an inner slice that contains this orbit, or by turning an outer face of the cube. To toggle the parity of a supercube "x-center" orbit you can only turn an outer face of the cube.
You have to look at all the center orbits to see which ones are odd parity and which are even parity. From this you can deduce which combination of inner slices quarter turns and/or outer face turns you must use to reduce all center orbits to even parity. Or, you can simply figure out which inner edge orbits are in an odd parity state, and the parity of the corners, and do the same thing. If all wing edge orbits are in an even permutation, and if corners are also in an even permutation, then every center orbit must be in an even permutation (on a supercube). This is a fundamental property of supercube centers.
Remember that supercubes are a little bit different from regular cubes, especially when it concerns centers.
On an even-dimensional cube, inner-slice turns do not affect the parity state of the centre pieces, while the parity state of the edges flips from odd to even for every inner-slice quarter turn. This means there is a 50% chance that the edges are in an odd parity state after solving the centres. After edge pairing, the odd parity state will manifest its self as a single pair of swapped edges (which appears as a single flipped edge pair after reduction).
Inner slices certainly change the parity of center orbits on even cubes, just not on the 4x4x4 supercube
That was the whole point of my proof above. Remember, you have to consider not how many 4-cycles occur because of an inner slice quarter turn, but how many 4-cycles
from which center orbits occur because of an inner slice quarter turn.
So I hope thats about right for OLL parity, but I'm still not certain about PLL parity. As Dan pointed out, its 2x edge 2-cycles. If this is the case, is PLL parity fixed at the end of edge pairing? Ie, would it be possible to go back to the edge pairing phase to fix PLL parity? Again, PLL parity doesn't happen on odd-dimensional cubes. Is this because the permutation of the centre edge pieces determines the overall permutation of the edges after reduction (ie, its not possible to swap only two centre-slice edges)?
PLL parity happens during edge pairing. When pairing the edges into groups you are also creating an edge cycle out of those groups. You have some control over where these edge groups end up, but we never think about this we just think about simply pairing them up. A good example of one case that is easy to see on 4x4. You have to pair the red-white edge group and the red-yellow edge group. Those pieces are in FR and FL on the cube. You can choose to pair them up using
d R F' U R' F d' or
u' R F' U R' F u. These two different algorithms will switch which pair ends up on which side of the cube (either FR or FL). During a speedsolve you have not figured out the parity of the corner orbit, or the resulting parity of the edge orbit based on where you can place the last two edge groups. However, if you were doing 4x4x4 FMC you could do this and figure out which alg to use such that the edge permutation you create after making those edge groups will match the corner permutation parity. This will avoid the PLL error or the PLL parity error.
Chris
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always no parity can give you some sub x0s avgs
