Prove that $T$ is compact

Question

$A \in B(X)$ and compact operator. Define a map $T: X/\mbox{ker}(A) \to X$. Where $T$ defined as

$T(x + \mbox{ker}(A)) = A(x)$ for all $x\in X$.

I want to prove that $T$ is compact.

I using the fact that $\mbox{ker}(A)$ is closed and if there exists and bounded sequence $(x_n)$ then $(A(x_n)$ has convergent sub-sequence, since $A$ is compact. But that doesn't help.

Answer 1

Hint: If $(x_n+ker (A))$ is norm bounded then there exists $(a_n) \subseteq A$ such that $(x_n+a_n)$ is norm bounded. Hence, $(T(x_n+ker (A))=(Ax_n)$ has a convergent subsequennce.

[By the definition of norm in $X/ker(A)$ we see that $\|x_n+\ker A\|+\frac 1 n > \|x_n+a_n\|$ for some $a_n \in A$].