Suppose that $F$ is a nonarchimedean local field, $G\_1$ and $G\_2$ are connected (linear) algebraic groups over $F$, and $\phi:G\_1\to G\_2$ is a surjective homomorphism of algebraic groups. Suppose $H$ is a hyperspecial maximal compact subgroup of $G\_1$. Is the image $\phi(H)$ necessarily a hyperspecial maximal compact subgroup of $G\_2$?



Image of a hyperspecial subgroup hyperspecial?
If by "surjective" you mean surjective in the usual sense (for example on $\overline{F}$-points) then maybe you have a problem, because $G\_1(F)$ may not surject onto $G\_2(F)$. So for example $SL(2)$ surjects onto $PGL(2)$ but if $R$ is the integers of $F$ then $SL(2,R)$ is hyperspecial max compact but its image in $PGL(2,F)$ isn't (it's not even maximal, as $PGL(2,R)$ strictly contains the image of $SL(2,R)$). 


However if $G\_1\to G\_2$ is, say, a $z$-extension, then (by definition) the kernel is central in $G\_1$ and has no $H^1$, so the long exact sequence shows $G\_1(F)\to G\_2(F)$ is surjective. Moreover, if I've got things right, then I think that $G\_1$ unramified forces the kernel to be unramified, and if you take a smooth integral model of $G\_1$ with $G\_1(R)$ equal to the hyperspecial you thought of, then the quotient of this model of $G\_1$ by the Zariski closure of the kernel will also be unramified, and the same cohomology argument shows that $G\_1(R)$ surjects onto $G\_2(R)$, so in this case you win.


