Given a covering space $(p,\tilde{X})$ of a space $X$, we know that every covering application $p:\tilde{X} \rightarrow X$ is a local homeomorphism and possesses the $\textbf{unique path lifting}$ property.
I was trying to think about an application $p:\tilde{X} \rightarrow X$, on some spaces $\tilde{X}$ and $X$, such that $p$ is a local homeomorphism and possesses the $\textbf{unique path lifting}$ property but was not a covering aplication (formally, $(p,\tilde{X})$ is not a covering space of X)
Some theorems of algebraic topology give some insight in this construction like: if $\tilde{X}$ is hausdorff we know that our application $p$ must not have $\#p^{-1}(x) = n \in \mathbb{N}$ $\forall x\in X$. This is, its fibers must not have the same cardinality, because that would make any local homeomorphism a covering application. Or it cannot be closed and have compact pre-images $p^{-1}(y)$ $ \forall x \in X$, for the same reason. (These are all characterizations of coverings of finite number of sheets.)
Is there an easy counter example for this?
We know that the composition of two covering spaces need not be a covering space. But in such a composition, local homeomorphism and path-lifting property stays valid. So consider any such example, which will give you a counter example.
The top map is a 2-fold covering but the bottom map has infinite fibers. The composition isn't locally trivial and is therefore not a covering map. However, it is still a semicovering map in the sense that it is a local homeomorphism which has the unique path lifting property.