Surjective (orthogonal) Projection

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Let T be a compact bounded operator on a Hilbert space H. Suppose that P is a (orthogonal) projection with $P \in B(H)$ and $P(H) \subseteq T(H)$. I am trying to show that $PT: H \to P(H)$ is surjective. Clearly $Image(PT) \subseteq P(H)$. So all we have to show is the reverse inclusion.

So far I have this: let $y \in P(H)$ be given. Then there exists an $x \in H$ so that $Px = y$. Since P is a (orthogonal) projection, $PPx = Px = y$. What I am inclined to try to show is that $PTx = PPx$, however I am not sure how to show this. So, am I on the right track with wanting to show $PTx = PPx$ or do I need to approach this another way? Any direction or help will be appreciated.

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We know that $P(H) \subseteq T(H)$. Thus, for every $z\in P(H)$ there exists $x,y\in H$ such that $Py=z= Tx$. Applying $P$ on both sides (and using that $P$ is a projection, i.e. $P^2 =P$) yiels $$ z= Py = PPy =PTx \in PT(H).$$ As $z\in P(H)$ was arbitrary, we get $P(H) \subseteq PT(H)$.