Let $G$ be a semisimple algebraic group over $Q_p$ and $K$ in $G(Q_p)$ a maximal compact open subgroup. Let $\tilde{\pi}\colon \tilde{G}\rightarrow G$ be the simply connected cover. Then $\tilde{\pi}^{1}(K)$ is a compact open subgroup of $\tilde{G}(Q_p)$. Is it necessarily maximal?
The answer is negative even for ${\rm{PGL}}_2$: the stabilizer of an edge in the building is a counterexample (with Iwahori preimage in ${\rm{SL}}_2(F)$). It likewise fails for ${\rm{PGL}}_n$ for every $n > 2$ by using stabilizers of chambers in the building.
In the work of BruhatTits, analyzing maximal compact open subgroups of $G(F)$ (for connected semisimple $G$ over a nonarchimedean local field $F$, say) is "easiest" when $G$ is split and simply connected. To understand more general split cases without the simply connected property involves additional work precisely to deal with the failure of surjectivity of $\widetilde{G}(F) \rightarrow G(F)$. To understand the origin of counterexamples with $G = {\rm{PGL}}_n$ (for $n > 1$), it is convenient to first carry out some general considerations (or jump ahead to the 2nd to last paragraph below if you are impatient).
Let $\pi: {\rm{SL}}_n \rightarrow {\rm{PGL}}_n$ be the natural central quotient map over $O_F$. The preimage in ${\rm{SL}}_n(F)$ of the (hyperspecial) maximal compact open subgroup $K_0 := {\rm{PGL}}_n(O_F)$ of ${\rm{PGL}}_n(F)$ is clearly $K'_0 := {\rm{SL}}_n(O_F)$, a (hyperspecial) maximal compact open subgroup in ${\rm{SL}}_n(F)$. I claim that $K_0$ is the unique maximal compact open subgroup of ${\rm{PGL}}_n(F)$ that contains $\pi(K'_0)$. To prove this directly, we briefly digress to recall an action construction.
The natural ${\rm{PGL}}_n$action on itself via conjugation uniquely lifts to an action of ${\rm{PGL}}_n = {\rm{GL}}_n/\mathbf{G}_m$ on ${\rm{SL}}_n$ in an evident manner, namely via ${\rm{GL}}_n$conjugation on ${\rm{SL}}_n$, say denoted $(g,x) \mapsto g.x$. Note that the action of $K'_0 = {\rm{PGL}}_n(O_F)$ on ${\rm{SL}}_n(F)$ preserves $K_0 = {\rm{SL}}_n(O_F)$. Thus, every ${\rm{PGL}}_n(F)$conjugate of $K_0$ has preimage in ${\rm{SL}}_n(F)$ that is a maximal compact open subgroup, with such preimages given by $g.K'_0$ for $g \in {\rm{PGL}}_n(F)$.
Now consider a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F) = {\rm{GL}}_n(F)/F^{\times}$ containing $\pi(K_0)$. We will show that $K \subset {\rm{PGL}}_n(O_F) = K'_0$, so $K = K'_0$ as desired. For any $k \in K$, consider the unique lift of $k$conjugation on ${\rm{PGL}}_n$ to an $F$automorphism $f_k$ of ${\rm{SL}}_n$. Concretely, this is conjugation in ${\rm{GL}}_n(F)$ against a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$.
Clearly $\pi(f_k(K'_0)) = k K_0 k^{1} \subset K$, so $f_k(K'_0) \subset \pi^{1}(K)$. But $\pi^{1}(K)$ is a compact open subgroup of ${\rm{SL}}_n(F)$ which contains the maximal compact open subgroup $K'_0$ and hence is equal to $K'_0$, so $f_k(K'_0) \subset K'_0$. Hence, a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$ normalizes $K'_0 = {\rm{SL}}_n(O_F)$; more precisely, $\widetilde{k} K'_0 \widetilde{k}^{1} = K'_0$ by volume considerations (using that ${\rm{SL}}_n$ is "unimodular" as an algebraic group over $F$). In terms of the standard $O_F$lattice $\Lambda = O_F^n$ inside $F^n$, this says that ${\rm{SL}}(\widetilde{k}.\Lambda) = {\rm{SL}}(\Lambda)$. But rather generally, a pair of $O_F$lattices $\Lambda_1$ and $\Lambda_2$ in $F^n$ satisfies ${\rm{SL}}(\Lambda_1) = {\rm{SL}}(\Lambda_2)$ inside ${\rm{GL}}(F^n) = {\rm{GL}}_n(F)$ if and only if $\Lambda_1$ and $\Lambda_2$ are $F^{\times}$multiples of each other, so $\widetilde{k} \in F^{\times} {\rm{GL}}(\Lambda) = F^{\times} {\rm{GL}}_n(O_F)$ and hence $k \in {\rm{PGL}}_n(O_F) = K_0$ as desired.
For $g \in {\rm{PGL}}_n(F)$, the ${\rm{SL}}_n(F)$conjugacy class of $g.K'_0$ only depends on the image of $g$ in the set $$\pi({\rm{SL}}_n(F))\backslash {\rm{PGL}}_n(F)/K_0 = F^{\times}/(F^{\times})^nO_F^{\times}$$ (use det). This set has size $n$, yet (as a special case of BruhatTits theory) ${\rm{SL}}_n(F)$ has $n$ distinct conjugacy classes of maximal compact open subgroups, so we have just built all of them. Hence, if $K$ is a maximal compact open subgroup of ${\rm{PGL}}_n(F)$ whose preimage $K'$ in ${\rm{SL}}_n(F)$ is a maximal compact open subgroup of ${\rm{SL}}_n(F)$ then necessarily $K' = g.K'_0$ for some $g \in {\rm{PGL}}_n(F)$. Thus, such $K$ would have to contain a conjugate of $\pi(K'_0)$. But we showed above that $\pi(K'_0)$ lies in a unique maximal compact open subgroup of ${\rm{PGL}}_n(F)$, namely $K_0$, so $K$ would have to be a conjugate of $K_0$!
The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$. Such $K$ exist for any $n \ge 2$, as stabilizers of a chamber in the building. But since the building can be most readily visualized and defined for $n = 2$, we now focus on describing an explicit counterexample for $n=2$.
Finally, consider ${\rm{PGL}}_2$. Let $K$ be the stabilizer of an edge in the building. Explicitly, this is described as follows. For the lattices $\Lambda = O_F^2$ and $\Lambda' = O_F \oplus \varpi O_F$ (with $\varpi$ a uniformizer of $F$) we choose some $g \in {\rm{GL}}_2(F)$ such that $g(\Lambda') = \Lambda$, so $g(\Lambda)$ corresponds to a line in the plane $\overline{\Lambda} := \varpi^{1}\Lambda/\Lambda$ over the residue field. By surjectivity of ${\rm{PGL}}_2(\Lambda) \rightarrow {\rm{PGL}}_2(\overline{\Lambda})$, we can adjust the choice of $g$ so that also $g(\Lambda) = \varpi^{1}O_F \oplus O_F = \varpi^{1}\Lambda'$. Then $g$ swaps the homothety classes of the lattices $\Lambda$ and $\Lambda'$, so $g^2$ preserves these homothety classes; i.e., $g^2$ lies in the compact open subgroup $$\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$$ that is the intersection of the stabilizers of the endpoints of an edge of the building whereas $g \not\in \mathbf{K}$. Clearly $$K = \mathbf{K} \coprod g \mathbf{K} = N_{{\rm{PGL}}_2(F)}(\mathbf{K}).$$ This is exactly the stabilizer of the midpoint of an edge in the building, and that midpoint is its unique fixed point. Since any compact subgroup of ${\rm{PGL}}_2(F)$ must fix some point of the building, it follows that $K$ must be maximal as a compact open subgroup of ${\rm{PGL}}_2(F)$.
Either for volume reasons or fixedpoint reasons with the building, $K$ is not conjugate to ${\rm{PGL}}_2(O_F)$, so its preimage in ${\rm{SL}}_2(F)$ is not maximal as a compact open subgroup of ${\rm{SL}}_2(F)$. Hence, by the general considerations above, $\pi^{1}(K)$ cannot be maximal as a compact open subgroup of ${\rm{SL}}_2(F)$, and this can also be seen by direct calculation: by determinant reasons, $$\pi^{1}(K) = \pi^{1}(\mathbf{K}) = {\rm{SL}}_2(O_F) \cap {\rm{SL}}(O_F \oplus \varpi O_F)$$ is the Iwahori subgroup given by the preimage in ${\rm{SL}}_2(O_F)$ of the upper triangular Borel subgroup of ${\rm{SL}}_2(k)$ (with $k$ the residue field of $O_F$).
It seems to me this is a completely general fact. If $\tilde\pi^{1}(K)$ is contained in a compact open subgroup $K'$, then $\tilde\pi(K')$ is a compact open subgroup containing $K$, thus equal to $K$; this implies $K'=\tilde\pi^{1}(K)$.

1$\begingroup$ I do not see why $\pi(K')$ should contain $K$, the map $\pi$ is not surjective on $Q_p$ points.... $\endgroup$ – user42721 Nov 23 '13 at 11:23