Remainders of compactifications are images of the Stone-Čech remainder. I need to show that if $\gamma X$ is a compactification of $X$, then $\gamma X\setminus X$ is the continuous image of $\beta X\setminus X$.  I know that there exists a continuous function from $\beta X$ onto $\gamma X$ that is the identity on $X$.  Furthermore, I know that this function can be defined as follows:
For $u\in\beta X$
if $u\in X$ let $f(u)=u$
else, let $f(u)\in \bigcap_{Z\in u} \text{cl}_{\gamma X} Z$
Another way of stating what I want to prove is, for $x\in X$, $f(p)=x$ implies $p=x$. To show this I think it would suffice to show $p$ (the ultrafilter) contains every zero set which contains $x$.
Here is my attempted proof...
Suppose $x\in X$ and $f(p)=x$, but $p\neq x$.  Then there is a zero set $Z$ with $x\in Z$ but $Z\notin p$.  Then (by an ultrafilter property) $Z\cap K=\phi$ for some zero set $K\in p$. 
I want to get a contradiction by showing these disjoint zero set do not have disjoint closures in $\beta X$.
$x\in \text{cl}_{\beta X}Z$ and $p\in \text{cl}_{\beta X}K$ so $f(p)=x\in f[\text{cl}_{\beta X}K]=\text{cl}_{\gamma X}K$
So $x\in \text{cl}_{\beta X}Z\cap \text{cl}_{\gamma X}K$.  But this is not quite what I need!
 A: One can actually prove a more general result without having to deal with the details of any construction of the Čech-Stone compactification:

Lemma. Let $X$ be a Hausdorff space and $f:X\to Y$ a continuous surjection. Let $D$ be a dense subset of $X$. If $f\upharpoonright D$ is a homeomorphism from $D$ to $f[D]$, then $f[X\setminus D]=Y\setminus f[D]$.
Proof. Suppose not, and let $p\in D$ and $q\in X\setminus D$ be such that $f(p)=f(q)=y$. Let $U_p$ and $U_q$ be disjoint open nbhds of $p$ and $q$, respectively. Then $U_p\cap D$ is an open nbhd of $p$ in $D$, so $f[U_p\cap D]$ is an open nbhd of $y$ in $f[D]$. Thus, $f[U_p\cap D]=V\cap f[D]$ for some open nbhd $V$ of $y$.
Let $W$ be any open nbhd of $q$; then $W\cap U_q\cap D\ne\varnothing$, so we may pick $x\in W\cap U_q\cap D$. Then $f(x)\in f[U_q\cap D]$, and $f[U_q\cap D]$ is disjoint from $f[U_p\cap D]=V\cap f[D]$ (since $f\upharpoonright D$ is a homeomorphism), so $f(x)\notin V$. That is, $f[W]\nsubseteq V$, and $f$ is not continuous at $q$, contrary to hypothesis. $\dashv$

In your setting $X$ is a dense subset of the Hausdorff space $\beta X$, and $f\upharpoonright X$ is the identity, which is certainly a homeomorphism.
