Two covering spaces covering each other are equivalent?

Question

Let $X_1,X_2,X$ be connected and locally path-connected spaces and $p_1:X_1\to X$ and $p_2:X_2\to X$ be covering maps.

Suppose that there are maps $f:X_1\to X_2$ and $g:X_2\to X_1$ such that $p_1=p_2f$ and $p_2=p_1g$.

My Question: Is it true that $p_1$ and $p_2$ are equivalent?

In other words, does there exists a homeomorphism $k:X_1\to X_2$ such that $p_1=p_2k$?

Thanks.

Answer 1

Yes, I think so. The existence of $f$ means $(p_1)_* \pi_1 X_1 \subseteq (p_2)_* \pi_1 X_2$ where $(p_1)_*$ and $(p_2)_*$ are just the inclusion of subgroups of $\pi_1 X$. Similarly, the existence of $g$ means $(p_2)_* \pi_1 X_2 \subseteq (p_1)_* \pi_1 X_1$. Therefore $\pi_1 X_1 = \pi_2 X_2$ as subgroups of $\pi_1 X$, and the classification of covering spaces implies $X_1$ and $X_2$ are equivalent covers.