Stinespring's theorem states the following:
Let $\mathfrak{A}$ be a unital $C^*$-algebra and $\Phi: \mathfrak{A} \rightarrow B(\mathcal{H})$ a completely positive map. Then there exists $\mathcal{K} \supset \mathcal{H}$ as subspace, a unital $*$-homomorphism $\pi : \mathfrak{A} \rightarrow B(\mathcal{K})$ and a bounded operator $V:\mathcal{H} \rightarrow \mathcal{K}$ such that
$$ \|V\|^2 = \|\phi(1)\| \quad, \quad \phi(a) = V^*\pi(a)V.$$
Why $V^*\pi(a)V$ can be seen as the 'compression' of $\pi(a)$? (I can't see it by just reading this written.. ). Does this fact hold only when $\Phi$ is unital, or does it always hold? (That $\pi$ can be seen as the 'dilation' of $\Phi$ or equivalently, $\Phi$ can be seen as the 'compression' of $\Phi$?
The usual proof of Stinespring's Theorem does not give you that $H\subset K$. What happens is that, when $\Phi$ is unital, you get $$1=\Phi(1)=V^*\pi(1)V=V^*V.$$ So $V$ is an isometry; in particular $V:H\to VH$ is a unitary. So we may replace $H$ with its image $VH$, and $\Phi$ with $\tilde\Phi:A\to B(VH)$ given by $\tilde \Phi(a)Vh=V\Phi(a)h$. The new map $\tilde\Phi$ is thus unitarily equivalent to $\Phi$, so it is the same map for most purposes. The map $VV^*$ is the orthogonal projection $P_{VH}$ of $K$ onto $VH$. So $$ \tilde\Phi(a)Vh=VV^*\pi(a)Vh=P_{VH}^{\vphantom{VH}}\pi(A)\,Vh. $$ If we forget the identifications, that is we think of $VH$ as $H$ and $\tilde\Phi$ and $\Phi$, then we have $H\subset K$ and $$ \Phi(a)=P_H\,\pi(a)|_H. $$ So $\Phi(a)$ is the 1,1 corner of $\pi(a)$. Arveson uses this point of view almost exclusively.