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In the case of surfaces, the necessary and sufficient condition is the absence of asymptotic values. A curve in $\gamma:[0,1)\to X$ is called an asymptotic curve if the limit $\gamma(t)$ as $t\to\infty$ is $\infty$ (the infinite point of one point compactification of $X$), but $f(\gamma(t))$ has a limit in $Y$ as $t\to 1$. This limit is called an asymptotic value. A covering is a local homeomorphism that has no asymptotic values. The only references for this that I know are for surfaces, but this should not be hard to prove in general under appropriate conditions. Probably $X$ and $Y$ have to be locally connected, in addition to what you said about them.

EDIT. Here is a proof. We assume that $Y$ is connected, and that every point $y\in Y$ has a path connected simply connected base of neighborhoods. Let $f:X\to Y$ be a local homeomorphism.

A path in a space is a continous map from $[a,b]$ or from $[a,b)$ to this space, where .

We say that the path begins at $\gamma(a)$. Let $\gamma:[a,b]\to Y$ be a path. A lifting of $\gamma$ is a path $\Gamma:[a,b]\to X$ such that $\gamma=f\circ\Gamma$.

If $\Gamma_1$ and $\Gamma_2$ are two liftings of the same path, and $\Gamma_1(a)=\Gamma_2(a)$ then $\Gamma_1=\Gamma_2$. (This is because $f$ is a local homeomorphism).

Let $x_0\in X$ be an arbitrary point and $y_0=f(x_0)$. Assuming that $f$ is a local homeomorphism without asymptotic values, we prove that every path beginning at $y_0$ has a lifting beginning at $x_0$. Let $\gamma$ be a path parametrized by $[a,b]$, beginning at $y_0$. Let $$c=\sup{ p>0:\gamma\vert_{[a,p]}\;\mbox{has a lifting beginning at}\; x_0}.$$ As $f$ is a local homeomorphism, we conclude that $c>a$. If $c=b$, the statement is proved. So assume that $c\in(a,b)$. Then, because of the uniqueness of lifting, a semi-open path $\gamma\vert_{[a,c)}$ has a lifting $\Gamma$.

I claim that $\Gamma(t)\to\infty$ as $t\to c$. Indeed, let $x_1\in X$ be a limit point of this path. Let $U$ be a neighborhood of $x_1$ where $f$ is a homeomorphism. Then $V=f(U)$ is a neighborhood of $\gamma(c)$, and we can extend our lifting by putting $\Gamma(t)=\phi(\gamma(t))$, for $t$ in a neighborhood of $c$, where $\phi$ is the inverse to the homeomorphism $f\vert_U$. This extends our lifting slightly beyond $c$, so we obtain a contradiction. Thus $\Gamma(t)\to\infty$ as $t\to c$. Then $\Gamma\vert_{[a,c)}$ is an asymptotic curve and $\gamma(c)$ is an asymptotic value. Again a contradiction.

Thus every path in $Y$ has a unique lifting, and this implies that $f$ is a covering.

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EDIT. Here is a proof.We assume that $Y$ is connected, and that every point $y\in Y$has a path connected simply connected base of neighborhoods.Let $f:X\to Y$ be a local homeomorphism.

A path in a space is a continous map from $[a,b]$ or from $[a,b)$to this space,where $a If$\Gamma_1$and$\Gamma_2$are two liftings of the same path,and$\Gamma_1(a)=\Gamma_2(a)$then$\Gamma_1=\Gamma_2$.(This is because$f$is a local homeomorphism). Let$x_0\in X$be an arbitrary point and$y_0=f(x_0)$.Assuming that$f$is a local homeomorphism without asymptotic values,we prove that every path beginning at$y_0$has a lifting beginningat$x_0$.Let$\gamma$be a path parametrized by$[a,b]$, beginning at$y_0$.$$c=\sup{ p>0:\gamma\vert_{[a,p]}\;\mbox{has a lifting beginning at}\;As$f$is a local homeomorphism, we conclude that$c>a$.If$c=b$, the statement is proved.So assume that$c\in(a,b)$. Then, because of the uniqueness of lifting,a semi-open path$\gamma\vert_{[a,c)}$has a lifting$\Gamma$. I claim that$\Gamma(t)\to\infty$as$t\to c$. Indeed, let$x_1\in X$be a limit point of this path. Let$U$be a neighborhood of$x_1$where$f$is a homeomorphism. Then$V=f(U)$is a neighborhood of$\gamma(c)$, and we can extend our lifting by putting$\Gamma(t)=\phi(\gamma(t))$, for$t$in a neighborhood of$c$,where$\phi$is the inverse to the homeomorphism$f\vert_U$.This extends our liftingslightly beyond$c$, so we obtain a contradiction.Thus$\Gamma(t)\to\infty$as$t\to c$. Then$\Gamma\vert_{[a,c)}$is an asymptotic curve and$\gamma(c)$is an asymptotic value.Again a contradiction. Thus every path in$Y$has a unique lifting,and this implies that$f$is a covering. 1 In the case of surfaces, the necessary and sufficient condition is the absence of asymptotic values. A curve in$\gamma:[0,1)\to X$is called an asymptotic curve if the limit$\gamma(t)$as$t\to\infty$is$\infty$(the infinite point of one point compactification of$X$), but$f(\gamma(t))$has a limit in$Y$as$t\to 1$. This limit is called an asymptotic value. A covering is a local homeomorphism that has no asymptotic values. The only references for this that I know are for surfaces, but this should not be hard to prove in general under appropriate conditions. Probably$X$and$Y\$ have to be locally connected, in addition to what you said about them.