What you will find, if you continue working the puzzle with the 6 there, is that it will force a contradiction later on.
Let's continue solving:
T= Top Row, M = Middle Row, B = Bottom Row
L= Left Column, C = Center Column, R = Right Column
- From the 4's in TC, MC, and BL, 4 must be in the bottom-left square of the BC.
- From the 2's in TC and MC, 2 must be in bottom right of BC.
- From the 9's in TL and TC, 9 must be in top right of TR
- From the 6's in TR and RC, 6 must be in bottom left of BL.
- From the 5 in TL, 5 must be in upper right of MC.
- There is now only one open space in the top row of the middle blocks, which must be a 7.
- From the 8 in LC, 8 must be in the middle right of BL.
- There is now only one open space in the right column of the left blocks, which must be a 1.
- From the 1's in ML and MR, 1 must be in middle left block of MC.
And NOW we run into a problem. We need an 8 somewhere in MC, but the only spaces available are in it's bottom row. But we can't put an 8 theethere because there's already an 8 in the bottom row of ML.