Let $X$ be a Hilbert space and $Y$ be a closed subspace, every vector $x \in X$ can be uniquely represented as $x=y+z$, $y \in Y$, $z \in Y^\bot$. Then the map
$$P_Y:X \to X, \quad P_Y(x)=y$$ is called the orthogonal projection in
$X$ onto $Y$.
I need to prove that $\text{Im}(P_Y)=Y$ and $\ker(P_Y)=Y^\bot$.
As for the image: the result seems to follow straight forward from the definition: $$P_Y(x)=y \in Y \quad \forall x \in X \implies \text{Im}(P_Y)=Y.$$
Kernel: it seems to be obvious but it's hard to think of any proof.
Best Answer
For the image you only proved one containment, namely ${\rm im}(P_Y)\subseteq Y$.
The other containment holds because $P_Y(y)=y$ for every $y\in Y$.
For the kernel, simply write the definitions: suppose an arbitrary $x\in X$ is given, decompose it as $x=y+z,\ y\in Y,\ z\in Y^\perp$, then $$P_Y(x)=P_Y(y+z)=y$$ So it's $0$ iff $y=0$ iff $x=z$ iff $x\in Y^\perp$.