TITLE: $X$ and $Y$ Banach Spaces, $T \in B(X,Y)$, $Y = \operatorname{im}T \oplus M$, for $M \subseteq Y$, then $\operatorname{im}T$ is closed in $Y$
QUESTION [5 upvotes]: Let $X$ and $Y$ be Banach spaces. If $T \in B(X,Y)$, and $Y = \operatorname{im}T \oplus M$ for some closed linear subspace $M$ of $Y$, then $\operatorname{im}(T)$ is closed in $Y$.

I am unsure if this statement is true or not. Nonetheless, I am having difficulty proving it, any suggestions or counterexample?

REPLY [4 votes]: It's true.
We can replace $X$ with $X/\ker T$ and $T$ with the induced map $\tilde{T} \colon X/\ker T \to Y$ if necessary, since $\operatorname{im} \tilde{T} = \operatorname{im} T$, and $X/\ker T$ is also a Banach space. Thus we may assume that $T$ is injective.
As a closed subspace of the Banach space $Y$, $M$ is itself a Banach space, and therefore $X \times M$ is also a Banach space if we endow it with one of the usual norms on a product of two normed spaces, e.g. $\lVert (x,m)\rVert = \lVert x\rVert_X + \lVert m\rVert_Y$.
We then define a continuous bijection $S \colon X \times M \to Y$ via
$$S(x,m) = Tx + m.$$
By the open mapping theorem, $S$ is open, and thus a homeomorphism. Since $X \times \{0\}$ is a closed subspace of $X\times M$, it follows that
$$\operatorname{im} T = S(X\times \{0\})$$
is a closed subspace of $Y$.