Show a function is continuous based on its properties relative to a covering map Let $p:E\to B$  be a covering map, let $Y$ be locally path-connected, and let $g:Y\to E$ be a function such that


*

*$p\circ g$ is continuous

*$g \circ \gamma $  is continuous for every path $\gamma$ in $Y$


Prove that $g$ is continuous.
I've been stuck on this problem for a while; some people have told me that the hypotheses that $Y$ be locally path-connected and that $g \circ \gamma $  is continuous for every path $\gamma$ in $Y$ are unnecessary.  I've proven a Lemma that says that if $y_0 $ and $y$ are path-connected in $Y$, there is a unique path in $E$ from $g(y_0)$ to $g(y)$ by using the uniqueness of path liftings in $B$.  I imagine this can be used to prove that $g$ is continuous, but I haven't made any progress on that front.
Any help is appreciated.
 A: You have to show that the inverse image $g^{-1}(U)$ of an open subset $U$ of $E$ is open.
Let $x\in U$,  there exists a neighborhood $V_x$ of $p(x)$ such that for every $y\in p^{-1}(p(x))$, there exists a neighborhood $U_y$ of $y$ such that $p_{\mid U_y}$ the restriction of $p$ to $U_y$ is an homeomorphism onto its image $V_x$, and $U_y\cap U_z$ is empty if $y\neq z$. $(p\circ g)^{-1}(V_x) =\cup_{y\in U_x}g^{-1}(U_y)$ is an open subset. Remark that $g^{-1}(U_y)\cap g^{-1}(U_z)$ is empty if $y\neq z$ since $U_y\cap U_z$ is empty.
Let see that $g^{-1}(U_y)$ is open for every $y\in p^{-1}(p(x))$. Let $z\in g^{-1}(U_y)$,  since  $z\in g^{-1}(V_x)$ there exists a neighborhood $W_z$ of $z$ contained in $g^{-1}(V_x)$. We can suppose that $W_z$ is connected by arcs, for every arc $c:I=[0,1]\rightarrow W_z$ such that $c(0)=z$, $g\circ c$ is an arc such that $g(c(0))\in U_y$, since $g\circ p$ is continuous, there exists $I'=[0,d]\subset I,d>0$ such that $(g\circ c)_{\mid I'}\subset U_y$ this implies that $g^{-1}(U_y)$ contains a neighborhood of $z$, thus is open. In particular, $g^{-1}(U_x)$ is open.
We can write $U=\cup_{x\in U} U_x$ such that $p_{\mid U_x}$ is an homoeomorphism onto its image, since $g^{-1}(U)=\cup_{x\in U}g^{-1}(U_x)$ is open since $g^{-1}(U_x)$ is open.
