Collision in Merkle–Damgård without a collision in compression function Is it possible to find a collision in Merkle–Damgård  just by omitting the extra one bit that is appended to each input without having a collision in compression function?
No, it is not possible.

If two inputs with different length produce identical hash outputs, then there would be a collision in the last invocation of the compression function. This is because the last input to the compression function always encodes the length of the original message.

If two inputs with identical length produce identical hash outputs, then there would be a collision in the compression function somewhere along the way. This is because considering the entire Merkle-Damgård hash as a hash tree build up of invocations of the compression function, then you will get two trees with identical structure, but those two trees have different inputs and identical output. By induction it can be shown that at least one of the invocations of the compression function in that tree structure must have a collision.