What is the mathematical property stating that it is hard to find a collision in the AES algorithm? Finding a collision in the AES algorithm is (apparently) hard. Why ? 

What is the mathematical property stating that there is a negligible probability to find a collision by taking two random plaintexts and setting the key ? 

Here I mean collisions like : $\operatorname{AES}_{K}(P_1)=\operatorname{AES}_{K}(P_2)$, with $P_1$ and $P_2$ being two random plaintexts and $K$ a 128 bits long fixed key.

In other terms, as AES 128 is mapping a 128 bits input to a 128 bits output, how can we be sure that all the outputs will be reached, for a given key ? What is the mathematical property in the design of AES making it a function "nearly" (fully ?) bijective ?

And, for AES-256, if we now set the plaintext and vary the keys (giving a function with input set size equal to $2^{256}$ and output set size equal to $2^{128}$), why is it difficult to find a collision ?

To clarify: for the last question only, I mean collisions like $\operatorname{AES}_{K1}(P_1)=\operatorname{AES}_{K_2}(P_1)$, with $P_1$ being a plaintext and $K_1, K_2$ two 256 bits long keys.
As mentioned in several of the other answers, AES is a family of permutations indexed by the key. Hence there cannot exist any collision of the form $AES_k(x) = AES_k(x')$ for $x\neq x'$. 

On the other hand, if we look at the function $f_x: k \rightarrow AES_k(x)$ we should ideally have a random function. Finding a non-random property of this function would be a major step in cryptanalysis of AES. So, with this reasoning the standard reasoning for random functions applies. For AES this means we have a function that maps 256 (192, or 128) bit strings to 128 bit strings. By the birthday paradox we know that an adversary that evaluates $f_x$ $q$ times has a success probability of $\mathcal{O}(\frac{q^2}{2^{128}})$ and that is also the best one can do without a huge improvement in the cryptanalysis of AES. However, this means one can find collisions of that function in about $2^{64}$ AES evaluations. This is already doable: the bitcoin network currently does about $2^{70}$ SHA2 computations per day!  And AES is faster than SHA2.