Compactness of operators in Banach spaces

Question

Let $X,Y,Z$ be Banach spaces. Let $T \in \mathbb L (X, Y)$ be a compact operator.

Why are the operators $B_1 \circ T \in \mathbb L(X,Z)$ and $T \circ B_2 \in \mathbb L(Z,Y)$ compact?

(Where $B_1 \in \mathbb L(Y,Z)$ and $B_2 \in \mathbb L(Z,X)$)

Answer 1

If $\{z_n\}$ is bounded in $Z$ then $\{B_2(z_n)\}$ is bounded in $X$ and hence $\{TB_2(z_n)\}$has a convergent subsequence. This proves that $T\circ B_2$ is compact. If $\{x_n\}$ is bounded in $X$ then $T(x_n)$ has convergent subsequence $T(x_{n_k})$. Just apply $B_1$ to this to see that $B_1\circ T$ is compact.