Does the Theory of one-to-one and not onto function is $\kappa$-categorical, for $\kappa > \aleph_0$ Language $L$ contains 1-arumgment function $f$ and relation symbol $=$.
The theory $\Sigma$ state that models must include a function $f$ which is one-to-one and not onto.
$\Sigma :\{ (\forall x)(\forall y)(f(x)=f(y)\rightarrow (x=y)) , (\exists z)(\forall y)((\exists x)f(x)=y \rightarrow \neg(y=z)) \}$
We were asked to find all cardinals $\kappa$ in which the theory $\Sigma$ is $\kappa$-categorical. I found that for finite and countable models $\Sigma$ is not $\kappa$-categorical. I think that the Theory is  $\kappa > \aleph_0 $ categorical but I hadn't succeeded proving it.
 A: (First of all there must be a mistake either in the formulas or in your description of them. As they stand, the first one says that if $f(x)=f(y)$, then $x\neq y$; and the second one says that $f$ is not onto, but has a unique point not attained. I'll answer the question using your description of the problem, and not the formulas)
The theory $\Sigma$ you decribe is not $\kappa$-categorical in any cardinality : it has no finite model, so we can not care about this; but it has two non isomorphic models of cardinality $\omega$: one in which $\omega\setminus \operatorname{Im}f $ is infinite and one where it is finite (actually there are many where it is finite).Actually, noticing this suffices to finish the proof, because this difference can be expressed in first order logic, and so these two models aren't even elementarily equivalent, which implies $\Sigma$ isn't complete, which implies it isn't $\kappa$-categorical for any infinite cardinal $\kappa$.
Indeed, letting $n$ be the number of points not in the image of the second model, this second model satisfies the sentence $\exists x_1,...\exists x_n, \forall y, (y\neq x_1)\land....\land (y\neq x_n) \implies\exists z, f(z)=y$, which isn't satisfied in the first model.
Note however, that you could do a similar proof that there are many non-isomorphic models of cardinality $\kappa$ for any infinite $\kappa$.
(As for the theory you've written down, it has no nonempty model, because if a model has a point $x$ then it satisfies $(x=x)\land f(x)=f(x)$. )
