Are models of computation closed under composition?
Given that, I would expect that for any reasonable model of computation, if $f : A \rightarrow B$ and $g : B \rightarrow C$ are computable, then $g \circ f : A \rightarrow C$ should be as well.


Let's say our model is quadratic time computation. If $f$ is the function which maps a string of length $n$ to a string of $n^2$ zeroes, then $f$ is computable in our model, but $f\circ f$ is clearly not.


  I'd expect that if $L_f = \{(a, b) \in A \times B : f(a) = b\}$ and $L_g = \{(b, c) \in B \times C : g(b) = c\}$ are decidable, then $L_{g \circ f} = \{(a, c) : \exists b \in B s.t. f(a) = b \wedge g(b) = c\}$ should be decidable as well.


This is a bit of a weird definition since $L_f$ can be much easier than $f$ (e.g., let $f(a)$ be the function that outputs a prime factorization of $a$). $F_l$ is poly-time while it is not known that $f$ is. Let's make $f$ also take an additional input $i$, and then let $f(a,i)$ output the number $i$ together with a factorization of $a$. If we then make $g$ output the $i$th bit of the factorization we end up with $L_f$ and $L_g$ being poly-time computable but $L_{f\circ g}$ being as hard as factorization.