Let $\tilde{G},G,H$ be topological groups with $H$ connected, $f:H\to G$ a continuous group homomorphism, and $p:\tilde{G}\to G$ be a covering map that is also a group homomorphism. Suppose $\tilde{f}:H\to \tilde{G}$ is a continuous lift of $f$ with $\tilde{f}(e)=e$. (Note that since $H$ is connected, such a lift $\tilde{f}$ is unique if it exists.) Then is it true that $\tilde{f}$ is necessarily a group homomorphism? At least we have $p\tilde{f}(xy)=f(xy)=f(x)f(y)=p\tilde{f}(x) p\tilde{f}(y)=p(\tilde{f}(x)\tilde{f}(y))$, so $\tilde{f}(xy)$ and $\tilde{f}(x)\tilde{f}(y)$ are contained in the same fiber of $p$, but I can't see why these two must be equal.
Lifting a group homomorphism via a covering map
algebraic-topologycovering-spacestopological-groups
Related Solutions
Strictly speaking, it doesn't make sense to say the monodromy action is equal to $\rho$ (even up to conjugation) as they are two actions on different sets. The correct notion of equivalence is isomorphism of group actions, defined as follows. If $G$ acts on both $X$ and $Y$, the two actions are isomorphic if there is a bijection $f: X \to Y$ such that $g\cdot f(x) = f(g\cdot x)$. You should convince yourself that this definition describes the notion of two actions being "the same".
If both sets are numbered, we sometimes conflate the actions with the isomorphic actions on $\{1,\ldots, d\}$, whereupon action isomorphism turns into conjugation. This can sometimes confuse matters, so let's be explicit: we are looking for a bijection between the fiber $\pi^{-1}(x)$ and $\{1,\ldots, d\}$ which commutes with the two actions of $\pi_1(V,q)$ (respectively, the monodromy action and the action $\rho$, given by $g\cdot x = \rho(g)(x)$).
The high-level explanation is that both the monodromy action and $\rho$ are isomorphic to the action of $\pi_1(V,q)$ on the cosets of $H$ (given by $b\cdot aH = (ba)H$).
The monodromy action is isomorphic to the action on cosets.
I will compose paths right to left, so that $ah \in aH$ is first $h$, then $a$. Then the lift of $ah$ based at $\tilde{q}$ (which I denote $\widetilde{ah}_{\tilde{q}})$ is the concatenation of $\tilde{h}_{\tilde{q}}$ (a loop in $\tilde{V}$) and $\tilde{a}_{\tilde{q}}$. In particular, the lift of every $ah \in aH$ based at $\tilde{q}$ has the same endpoint, namely $\tilde{a}_{\tilde{q}}(1) \in \pi^{-1}(q)$. This gives a function $f:G/H \to \pi^{-1}(q)$ by $f(aH) = \tilde{a}_{\tilde{q}}(1)$. Any $y \in \pi^{-1}(q)$ can be connected to $\tilde{q}$ by a path $\tilde{\gamma}$, and $f(\pi(\tilde{\gamma}) H) = y$, so $f$ is moreover bijective.
$b \in \pi_1(V,q)$ acts on $x \in \pi^{-1}(q)$ by sending it to $\tilde{b}_x(1)$. When $x = f(aH) = \tilde{a}_{\tilde{q}}(1)$, we get $$b\cdot f(aH) =b\cdot \tilde{a}_{\tilde{q}}(1) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1),$$ which is a complicated way of saying "first lift $a$ based at $q$, then lift $b$ based at the endpoint of $\tilde{a}_{\tilde{q}}$, and take the endpoint of that". But by uniqueness of lifts, this is the same as just taking the endpoint of $\widetilde{ba}_{\tilde{q}}$. Then $$b\cdot f(aH) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1) = f((ba)H) = f(b\cdot aH).$$ Then the two actions are isomorphic, as claimed.
The action on cosets is isomorphic to $\rho$.
The final piece of the puzzle is pure group theory—if $\rho: G \to S_n$ is a transitive* action (via $g\cdot x = \rho(g)(x)$), then it is isomorphic to the action on the cosets of the stabilizer $H = \text{Stab}(1)$. The isomorphism is given by $j(aH) = a \cdot 1$, and I'll leave it to you to verify it is indeed an isomorphism of actions (it's essentially a much easier version of the above).
Then $j\circ f^{-1}: \pi^{-1}(q) \to \{1,\ldots,d\}$ is an isomorphism between the monodromy action of $\tilde{V}$ and $\rho$.
*If $\rho$ is not transitive, then the statement is false, as the monodromy action of a connected covering space is always transitive.
$\newcommand{\wt}{\widetilde}\newcommand{\by}{\times}\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}\newcommand\Q{\mathbb Q}$
The $\wt f$ you describe will not always be continuous. As a simple case, say $G=H=\R$ and let $\wt G=\R\by\Q$ with covering map given by projection onto $\R$. Let $f:\R\to\R$ be the identity, so choosing a lift $\wt f:\R\to\R\by\Q$ amounts to choosing a group homomorphism $\R\to\Q$. There are plenty of these. For example, observe that $\R$ is a $\Q$-vector space and fix an (uncountable) basis $\{e_i\}_{i\in I}$. Let $\lambda_i:\R\to\Q$ be projection onto the $i$th factor, i.e. given $r\in\R$, one has $$r=\sum_{i\in I}\lambda_i(r)e_i.$$ Then, $\wt f(r):=(r,\lambda_i(r))$ is a lift of $f$ and a group homomorphism for all $i\in I$, but these maps are typically not continuous.
Best Answer
The trick is you can use the uniqueness of lifts to compare $\tilde{f}(xy)$ and $\tilde{f}(x)\tilde{f}(y)$. Let $g(y)=\tilde{f}(xy)$ and $h(y)=\tilde{f}(x)\tilde{f}(y)$. Then $g$ and $h$ are both lifts of $pg=ph:H\to G$ with $g(e)=h(e)=\tilde{f}(x)$. Therefore, they are equal.