Reversing N CRC steps Starting from a known CRC result(let's call it CRC1), could I undo the CRC operations for last N bits, so I could obtain the CRC (let's call it CRC2) of the message sequence without the last N bits. 
$$
M(x) - message \\ M = \{M1,LastBits\} \\CRC_1=CRC\{M\}\\CRC_2=CRC\{M_1\}
$$
Does a function F() exist for which:
$$F(CRC_1,LastBits) = CRC_2$$

My hunch would be that such a function which would unrewind the last N CRC steps, would have 2N possible solutions.
This is easiest to understand if we use polynomial arithmetic.

The CRC of a message $m(x)$ is the remainder $r(x)$ of $m(x) x^k$ when divided by the CRC polynomial $f(x)$. Or more conveniently, the CRC is congruent to the message multiplied by $x^k$ modulo the CRC polynomial, $r(x) \equiv m(x) x^k \pmod{f(x)}$.

If the message consists of a prefix $m_1$ and a suffix $m_2$ of length $n$, we can express that as $m(x) = m_1(x) x^n + m_2(x)$. If the CRC of $m_1$ is $r_1(x)$ and the CRC of $m_2(x)$ is $r_2(x)$, then the CRC of $m(x)$ is $$r(x) \equiv (m_1(x) x^n + m_2(x)) x^k \equiv r_1(x) x^n + r_2(x) \pmod{f(x)}.$$

If $f(x)$ is not a multiple of $x$ (which it won't be), there is a polynomial $g(x)$ such that $x g(x) \equiv 1 \pmod{f(x)}$, and then $$r_1(x) \equiv r_1(x) (x g(x))^n \equiv (r(x) - r_2(x)) g(x)^n \equiv (r(x) - m_2(x) x^k) g(x)^n \pmod{f(x)}.$$

In other words, you find what you ask for by first finding $g(x)$, then computing a difference, multiplying by a suitable power of $g(x)$, then taking the remainder when dividing by $f(x)$.