Suppose $p: Y \to X$ and $p': Y' \to X$ are covering maps, and let $\phi: Y \to Y'$ be a homeomorphism such that $p'\phi=p$. Show that the functors $p^{-1}$ and $(p')^{-1}$, from $\Pi_1(X)$ to the category of sets, are naturally isomorphic. I changed the notation from $p'=p_2$ and $p = p_1$
$\textbf{ My Attempt:}$ We are given that $\phi$ is a homeomorphism from $Y \to Y'$, thus $\phi_c : p^{-1}(c) \to (p')^{-1}(c)$ will be an isomorphism for any $c \in Ob(\Pi_1(X))$. \newline
We must show that $$\phi_{c'} \circ p^{-1}(f)= (p')^{-1}(f) \circ \phi_c$$
We have $$p_1^{-1}(x) \to \phi(p_2^{-1}(x)) \to \phi(p_2^{-1}(x)) * f $$ defined by $$ c \to \phi_c \to \phi_c * f$$ Thus $$p_1^{-1}(x) \to p_1^{-1}(x) * f \to \phi( p_1^{-1}(x) * f) $$ defined by $$ c \to c * f \to \phi_{c * f}$$
Where $\phi_{c} * f$ represents the endpoint of a path starting at $\phi_{c}$ and concatenated with $f$.
We want to show that $\phi_{c*f} = \phi_c * f$ The values of $\phi_c * f$ are uniquely determined by $f$ and $p_2( \phi_c * f) = x $ for some $x \in X$, such that $ p(c) * f = x$
Thus $p_2 ( \phi_c * f)=x$ and we also have that $p_1 ( c * f) =x$. Notice that $$p_2 ( \phi_{c*f})=p_1(c*f)=x$$
Thus $$p_2 ( \phi_{c} * f)= p_2 ( \phi_{c * f})=p_1( c * f)=x$$ Thus since $f$ is unique we have that $ \phi_{c}* f = \phi_{c *f}$
I am so stuck on the very last sentence. I am sure that the uniqueness of $f$ solves the problem but I am not sure how to explicitly show it. Any help is greatly appreciated!!
Look at the starting point of the path $\phi_{c*f}$, since $f$ is the homotopy lift from $X$ starting at $x \in X$, we have $f(0)=c$. It follows that $\phi_{c*f_0} = \phi_c$ and so the starting point of $\phi_{c*f}$ is $\phi_c$. Unique path lifting then gives the result since we know there exists only one unique path in $Y'$ that starts at $\phi_c$ and ends as described above. Thus $\phi_{c} * f=\phi_{c * f}$