Why is this operator one-to-one I am reading a textbook, and would like to ask a question about the proof. 

Here $S_p$ is the Schatten p class. 
My question is, in the proof, why is $A: X\to H$ is one-to-one? I actually don't know what the notation $\ominus$ mean (there is no explanation in the context).  
 A: The notation $X= H \ominus \ker(A)$ means that $H = X \oplus \ker(A)$. Here $\oplus$ is the direct sum of the vector spaces $X$ and $\ker(A)$. Every element $h$ in $H$ can be written (in a unique way) as 
$$
h=x +k,
$$
where $x \in X$ and $k \in \ker(A)$.
The notation $\ominus$ is then meant to suggest subtraction of the subspace $\ker(A)$:
$$
H \ominus \ker(A) = (X \oplus \ker(A)) \ominus \ker(A) = X.
$$
That $H$ is indeed equal to $X \oplus \ker(A)$ is the isomorphism theorem for vector spaces (http://en.wikipedia.org/wiki/Isomorphism_theorem).
It is important to note that $\ker(A) \cap X =\{0\}$. So the restriction of $A$ to $X$ will have kernel equal to $\{0\}$. In other words, this restriction is one-to-one. You are told in the question that $A$ is surjective. Its range will not have been altered by restricting its domain to $X$ because
$$A(h)=A(x+k)=A(x)+A(k)=A(x)+0=A(x).$$
Also, as Freeze_S notes in his comment, $\ominus$ denotes the orthogonal difference $H \ominus \ker(A)=\ker(A)^{\perp}$. As $A$ is continuous, $\ker(A)$ is closed and this leads to the same decomposition described above.
