It is a well-known result about covering spaces that if $f_i:X_i\to Y $ $ (i=1,2)$ are coverings of $Y$ and $\sigma:X_1\to X_2 $ is a morphism of covbering spaces, then $(X_1,\sigma ) $ is a covering of $ X_2$. Morover, the degree of $f_1:X_1\to Y $ is the degree of $\sigma:X_1\to X_2 $ times the degree of $f_2:X_2\to Y$. From this it follows that a morphism of covering spaces of the same degree must be a biyective continuous function. My question is, under which hypothesis can I know for certain that the inverse is continuous too? Thank you all in advance
Pd: Everything here has finite fibres and the spaces are all path-connected.
The inverse is always continuous since since a covering is a local homeomorphism; i.e if $f:X\rightarrow Y$ is a covering, $x\in X$, there exists neighborhoods $U$ of $x$, $V$ of $p(x)$ such that $p(U)=V$ and the restriction of $p$ to $U$ is a diffeomorphism onto its image.