Why is image $u^\ast$ closed Let $H,H'$ be Hilbert spaces and $u\in B(H,H')$. Let $u$ be Fredholm. Then there exists a pseudo inverse $v\in B(H',H)$ such that $u = u v u$ and $v$ is Fredholm. Then $u^\ast = u^\ast v^\ast u^\ast$ hence $(u^\ast v^\ast)^2 = u^\ast v^\ast$.
The proof continues as follows:
$u^\ast (H') =u^\ast v^\ast(H)$ hence $u^\ast (H')$ is closed.

(1) Why is $u^\ast (H') =u^\ast v^\ast(H)$?

and 

(2) Why does this imply that $u^\ast (H')$ is closed?

 A: To see (1), we must show that $\operatorname{im} (u^\ast\circ v^\ast) \subset \operatorname{im} u^\ast$, and $\operatorname{im} u^\ast \subset \operatorname{im}(u^\ast\circ v^\ast)$.
For any (linear) operators $S,T$ such that $S\circ T$ is defined, we have $\operatorname{im} (S\circ T) \subset \operatorname{im} S$. That shows the first inclusion, setting $S = u^\ast$ and $T = v^\ast$. Then we use the identity $u^\ast\circ v^\ast \circ u^\ast$ and set $S = u^\ast \circ v^\ast$ and $T = u^\ast$ to obtain
$$\operatorname{im} u^\ast = \operatorname{im} \bigl((u^\ast \circ v^\ast) \circ u^\ast\bigr) \subset \operatorname{im} (u^\ast \circ v^\ast),$$
and together $\operatorname{im} u^\ast = \operatorname{im} (u^\ast\circ v^\ast)$.
To see (2), note that $P = u^\ast \circ v^\ast$ is a continuous projection, hence $Q = I-P$ also is a continuous projection - $Q^2 = I - 2P + P^2 = I - 2P + P = I-P = Q$ - and hence $\ker Q$ is closed. But $\ker Q = \operatorname{im} P$. So the range of a continuous projection is always closed, in particular $(u^\ast \circ v^\ast)(H) = u^\ast(H')$.
