Subspace isomorphic to a complemented subspace

I'll begin by writing down the definitions I'm using, to avoid confusion.

Let $X$ be a Banach space and let $Y$ be a subspace of $X$. We say that $Y$ is complemented in $X$ if there exists a linear continuous operator $P : X \to Y$ such that Im$(P) = Y$ and $P \circ P =P$. Equivalently, $Y$ is complemented if $Y$ is closed and there is a closed subspace $W$ of $X$ such that $X = Y \oplus W$.

My question is this: Let $X$ be a Banach space, $Y$ a complemented subspace of $X$ and $Z$ a (closed) subspace of $X$ isomorphic to $Y$. Must $Z$ be complemented in $X$ as well?

The natural idea would be to take $P : X \to Y$ a projection and $T : Y \to Z$ an isomorphism, and define $Q = T \circ P : X \to Z$. The problem I'm having is showing that $Q(z)=z$, for all $z \in Z$, or $Q^2(x)=Q(x)$, or all $x \in X$. It's enough that $P(z)=T^{-1}(z)$, but I don't know if I can guarantee this - and this might be the clue to finding a counterexample.

A simple counterexample is $X=\ell_\infty\oplus c_0$. The second summand, $c_0$, is complemented in $X$. But the copy of $c_0$ sitting in $\ell_\infty$ is not complemented in $\ell_\infty$, hence not complemented in $X$.
• I don't understand your argument. So, if $S \subset \ell_\infty \oplus c_0$ was a complement for $\{0\} \oplus c_0 \subset c_0 \oplus \ell_\infty$, are you suggesting that $\pi_2(S)$ would be a complement for $c_0 \subset \ell_\infty$ where $\pi_2$ is the factor projection $c_0 \oplus \ell_\infty$? I don't even see why $\pi_2(S)$ would be closed.... – Mike F Nov 18 '14 at 1:30
• Indeed, if $T : X \to Y$ is a closed operator, then graph $\Gamma \subset X \times Y$ of $T$ would have $\pi_2(\Gamma) = ran(T)$, which may or may not be closed in $Y$. – Mike F Nov 18 '14 at 1:33
• @MikeF No. It's easiest to see using the projection, I think. If there were a continuous projection $P \colon X \to X$ with image $c_0 \oplus \{0\}$, the restriction to $\ell_{\infty} \oplus \{0\}$ would induce a continuous projection $Q \colon \ell_{\infty} \to \ell_{\infty}$ with image $c_0$. Thus $c_0$ would be complemented in $\ell_{\infty}$ (which it isn't, so $P$ doesn't exist). – Daniel Fischer Aug 12 '18 at 12:47