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A covering map is a continuous surjection $p:\tilde{X}\to X$ for which each point has an open neighbourhood $U$ with $p^{-1}(U)$ a disjoint union of open sets, each mapped homeomorphically onto $U$. A connected simply connected covering space is a universal cover.

There is a necessary connectedness qualification in the printed conclusion. Let $A_0$ be the path component of $a_0$ in $A$. Then $p(\tilde{A})=A_0$: one inclusion follows by projecting paths, and the other by lifting a path in $A_0$ from $a_0$. Thus the asserted surjection onto all of $A$ requires $A$ to be path connected. Without it, take $p$ to be the identity of $\mathbb{R}$ and $A$ to be two disjoint intervals; a selected component cannot cover both.

To prove the assertion about a component of a pulled-back universal cover, choose an evenly covered neighbourhood $U$ in $X$. Local path connectedness of $A$ supplies a path-connected relatively open neighbourhood $V\subseteq U\cap A_0$. Each inverse sheet over $V$ is path connected, so it lies wholly in $\tilde{A}$ or is disjoint from it. These sheets evenly cover $V$ under $p{|}_{\tilde{A}}$. Therefore $p{|}_{\tilde{A}}:\tilde{A}\to A_0$ is a covering map.

Under the classification of connected covering spaces, its subgroup consists of based loops in $A_0$ whose lifts close at ${\tilde{a}}_0$. A loop viewed in $X$ closes in the universal cover exactly when its class in ${\pi }_1(X,a_0)$ is trivial. Hence the subgroup isHere ${\pi }_1(A,a_0)={\pi }_1(A_0,a_0)$, so the subgroup formula itself needs no change. Injectivity of the covering's induced fundamental-group map is the usual homotopy-lifting property.