How is the affine cipher a permutation? It seems there would be some collisions in what the different symbols map to, resulting in loss of information. 

For example, say we use the English alphabet, multiply by 2 and take the congruence mod 26.  In this case, A and N will both map to the same ciphertext letter, regardless of what the shift value is.  Same with B and O, etc.  In fact, if the formula is $(ax + b) \mod n$ where $n$ is the size of the plaintext alphabet, then the ciphertext alphabet will have only $1 \over a$ as many symbols as the plaintext alphabet.  

Am I misunderstanding something?
Multiplication by 2 modulo 26 is indeed not a permutation.  For example, $$2 \times 13 = 26 \equiv 0 = 2 \times 0 \pmod{26}.$$

The necessary and sufficient criterion for the affine map $x \mapsto ax + b$ to be a permutation of the integers modulo $m$ is that the multiplicative constant $a$ and the modulus $m$ must be coprime; that is, their greatest common divisor must be 1.

If this is the case, then we can find the modular multiplicative inverse $a^{-1}$ of $a$ modulo $m$ and use that to define the inverse map $x \mapsto a^{-1}(x-b)$.  It's easy to verify, based on the fact that $a^{-1} a \equiv 1 \pmod m$, that the composition of these two maps (in either order) yields the identity map $x  \mapsto x$, which is sufficient to show that these maps must be bijective.

Conversely, if $a$ and $m$ share a common divisor $g &gt; 1$, then the map $x  \mapsto ax + b$ cannot be a bijection.  In particular, both $0$ and $m/g$ will be mapped to the same element $b$ by this map.