Condition for a function $\widetilde \phi:\mathbb{H}\rightarrow \mathbb{H}$ to descend to $S_g$ Let $S_g$ be a closed and oriented surface of genus $g\ge 2$ and fix any universal cover $\pi:\mathbb{H}\rightarrow S_g$. The group $\pi_1(X)$ acts on $\mathbb{H}$ through the monodromy action. 
Given $\widetilde x\in \mathbb{H}$ and $[\gamma]\in \pi_1(X)$ then $[\gamma]\cdot \widetilde x=\widetilde \gamma(1)\in \mathbb{H}$: one first considers any representative $\gamma$ of $[\gamma]$ with base point in $\pi(\widetilde x)$, then $\widetilde \gamma$ is the lifting of $\gamma$ to $\mathbb{H}$ such that $\widetilde \gamma(0)=\widetilde x$.
I am trying to completely understand universal covers, liftings and homotopy, so I would like to know if my following statements are true:

It is my understanding that any continuous function $\widetilde \phi:\mathbb{H}\rightarrow \mathbb{H}$ descends to a function $ \phi:S_g\rightarrow S_g$ if there is an endomorphism $\phi_*:\pi_1(X)\rightarrow \pi_1(X)$ such that it results $\widetilde\phi([\gamma]\cdot x)=\phi_*([\gamma])\cdot \widetilde\phi(x)$. Then $\phi_*$ is the map induced by $\phi$ on $\pi_1(X)$.
In particular any continuous function $\widetilde f:\mathbb{H}\rightarrow \mathbb{H}$ such that $\widetilde f([\gamma]\cdot x)=[\gamma]\cdot \widetilde f(x)$ for every $x\in \mathbb{H}$ and $[\gamma]\in \pi_1(X)$ descends to a function $f:S_g\rightarrow S_g$ such that $f_*=Id$.

This is my proof: given any continuous function $\psi:S_g\rightarrow S_g$ one has the following commuting diagram
$$\begin{array}
$\mathbb{H} & \stackrel{\widetilde \psi}{\longrightarrow} & \mathbb{H}\\
\downarrow{\pi} & & \downarrow{\pi} \\
S_g & \stackrel{\psi}{\longrightarrow} & S_g  
\end{array}
$$
and consequently one has $\psi(\gamma)=\pi(\widetilde \psi(\widetilde \gamma))$ (where as before $\widetilde \gamma$ is the lifting such that $\widetilde \gamma(0)=\widetilde x$). This implies 
$$\widetilde{\psi(\gamma)}(1)=\widetilde \psi(\widetilde \gamma(1))$$
where $\widetilde{\psi(\gamma)}$ is the lifting of $\psi(\gamma)$ such that $\widetilde{\psi(\gamma)}(0)=\widetilde \psi(\widetilde x)$. Finally, one notes $\widetilde \psi(\widetilde \gamma(1))=\widetilde \psi(\gamma\cdot \widetilde x)$ and $\widetilde{\psi(\gamma)}(1)=[\psi(\gamma)]\cdot \widetilde \psi(\widetilde x)=\psi_*[(\gamma)]\cdot \widetilde \psi(\widetilde x)$.
Consequently, any continuous function $\widetilde \phi:\mathbb{H}\rightarrow \mathbb{H}$ with such property will descend to $S_g$.
Is my reasoning correct? If not, which additional conditions must a continuous function $\widetilde \phi:\mathbb{H}\rightarrow \mathbb{H}$ satisfy in order to descend to a function $\phi:S_g\rightarrow S_g$? And in particular to a function such that $\phi_*=Id$?
 A: First of all, your definition of the action of $\pi_1$ on $\mathbb{H}$ is wrong and much of what you say does not make sense because you have thoroughly neglected basepoints.  There is no such thing as $\pi_1(S_g)$.  There is only $\pi_1(S_g,x)$ for any chosen basepoint $x\in S_g$.
Here is a correct way to state everything.  None of this is special to surfaces, so I will consider arbitrary spaces.  Let $X$ be a path-connected space with $\pi:Y\to X$ a universal cover.  Fix a basepoint $x_0\in X$ and a basepoint $y_0\in\pi^{-1}(x_0)$.  Then $\pi_1(X,x_0)$ acts on $Y$ as follows.  Given $[\gamma]\in\pi_1(X,x_0)$ and $y\in Y$, choose a path $\delta$ from $y_0$ to $y$ in $Y$.  We can then concatenate $\pi\delta$ and $\gamma$ to get a path $\gamma*\pi\delta$ from $x_0$ to $\pi(y)$ in $x$.  Let $\epsilon$ be the lift of $\gamma*\pi\delta$ to a path in $Y$ starting at $y_0$.  We then define $[\gamma]\cdot z=\epsilon(1)$.
As for the statement you are asking to be true, it does not even make sense, again because of basepoints: a map $\phi:X\to X$ does not induce an endomorphism of $\pi_1(X)$.  Instead, it induces a homomorphism from $\pi_1(X,x_0)\to \pi_1(X,\phi(x_0))$ for any choice of basepoint $x_0$.  You only get an endomorphism if $\phi(x_0)=x_0$.
You also seem to be conflating two quite different questions: when a map $\phi:S_g\to S_g$ exists and what you can say about the induced map on $\pi_1$.  You have given an argument for the second question but not the first question, but you have phrased much of your post as though the first question is your primary interest.  In fact, the first question is rather trivial.  Since any covering map is a quotient map, the sole condition needed for a map $\widetilde{\phi}:Y\to Y$ to give a map $\phi:X\to X$ is that for all $y,y'\in Y$, $\pi(y)=\pi(y')$ implies $\pi(\widetilde{\phi}(y))=\pi(\widetilde{\phi}(y'))$.  This implies you get a well-defined map of sets $\phi:X\to X$, which is then automatically continuous since $\pi$ is a quotient map.
As for the action on $\pi_1$, your statement and proof are correct if you additionally assume that $\phi(x)=x$ and are talking about the fundamental group with $x$ as the basepoint.  More generally, if you take $\phi_*$ to instead be a homomorphism $\pi_1(X,x)\to\pi_1(X,x')$ and $\phi(x)=x'$, the argument still works to show that $\phi_*$ is the induced map on $\pi_1$.
