# Why is the image of closed invariant subsets closed? Mumford, GIT, Theorem 1.1

I'm currently studying Mumford's Geometric Invariant Theory. Unfortunately, I'm stuck understanding a detail in Theorem 1.1.

## (Partial) Claim of Theorem 1.1

Let $$X = \operatorname{Spec} R$$ be an affine scheme over a characteristic-zero field $$k$$ and consider a reductive group action $$G \curvearrowright X$$. Then there is a categorical quotient of $$X$$ by $$G$$, induced by the inclusion $$\phi\colon R_0 \hookrightarrow R$$ ($$R_0$$ being the ring of invariants).

## Part where I'm stuck

Now, he wants to use Remark 6 on page 8 to derive the claim.

### Remark 6 on page 8

Let $$G$$ be a group scheme acting on a scheme $$X$$ (via $$\sigma\colon G\times X \to X$$). Furthermore, let $$\phi\colon X \to Y$$ be a morphism of schemes. The following conditions together imply that $$(Y, \phi)$$ is a categorical quotient of $$G\curvearrowright X$$:

1. $$\phi\circ\sigma = \phi\circ p_2$$ (where $$p_2$$ is the projection to the second component of the cartesian product)
2. $$\phi$$ induces an injective morphism of schemes $$\mathcal{O}_y \hookrightarrow \phi_*\mathcal{O}_X$$ that has the subsheaf of invariants as its image.
3. If $$W$$ is an invariant closed subset of $$X$$, then $$\phi(W)$$ is closed in $$Y$$; if $$(W_i)$$ is a family of invariant closed subsets of $$X$$, then $$\phi\left(\bigcap_i W_i\right) = \bigcap_i \phi(W_i)$$.

### Mumford's proof sketch for condition (3)

I understand how Mumford proves the first two conditions, but I don't get the third one. Mumford shows that for closed invariant sets $$(W_i)$$,

$$\overline{\phi\left(\bigcap_i W_i\right)} = \bigcap_i \overline{\phi(W_i)}.$$

Now, it suffices to prove that $$\phi(W_1)$$ is closed.

Mumford then prompts me to apply the above equation to the case where $$W_1$$ is arbitrary and $$W_2$$ is the preimage of a closed point of $$Y$$, and claims that this implies that $$\phi(W_1)$$ is closed.

This last claim is the one I don't understand.

## My thoughts

• In the situation where $$W_2$$ is the preimage of a closed point $$y \in Y$$, both sides of the equation we already have are subsets of $$\phi(\phi^{-1}(y))$$. If this set is empty, the equation just states $$\varnothing = \varnothing$$, not very helpful.
• Since the left hand side is either $$\overline{\{y\}}$$ or $$\overline{\varnothing}$$, the closure on the left-hand side is irrelevant in the situation of the previous bullet point.
• Let's assume that $$\phi$$ is surjective. Then the right-hand side is $$\{y\}$$ if and only if $$y\in \overline{\phi(W_1)}$$ and $$\varnothing$$ otherwise. So if $$y\in \overline{\phi(W_1)}$$, the equation implies $$y\in \bigcap_i \phi(W_i) \subseteq \phi(W_1)$$. $$\phi(W_1)$$ contains all closed points of its closure. However, what does this help, and is $$\phi$$ really surjective?

Hints for how to expand Mumford's proof sketch are highly appreciated.

The idea is actually rather simple. Let $$W$$ be a closed $$G$$-invariant subset of $$X$$, and $$y$$ a closed point that is not in $$\phi(W)$$. Note that $$\phi^{-1}(y)$$ is also closed and $$G$$-invariant. We already know that $$\overline{\phi(W\cap\phi^{-1}(y))}=\overline{\phi(W)}\cap\{y\}.$$ But the LHS is empty which means that $$y\notin\overline{\phi(W)}$$. This tells us that $$\phi(W)$$ must be closed. Otherwise, we may find a closed point $$y\in\overline{\phi(W)}\setminus\phi(W)$$ that is not contained in $$\overline{\phi(W)}$$ by the argument above which is absurd. See the proof of Theorem 6.1 in Dolgachev's book "Lectures on Invariant Theory" for more details.
FYI, the morphism $$\phi$$ is submersive which means that it is surjective and the induced topology on its image is the quotient topology. One could show that any categorical quotient satisfying the 3 conditions in remark 6 is submersive.