I am stuck with the following simple (i guess) but technical problem.

Let $G$ be a complex reductive group, $\mathcal{O}:=\mathbb{C}[[t]]$ the formal power series, $\mathcal{K}=\mathbb{C}((t))$ the formal Laurant series, and $P^+$ the set of dominant integral (co)-weights of $G$. Then there exists a Bruhat decomposition $$G(\mathcal{K})=\bigsqcup_{\lambda} G(\mathcal{O})\lambda G(\mathcal{O})$$ Let furher denote $G(\mathcal{K})^\lambda:= G(\mathcal{O})\lambda G(\mathcal{O})$ and $\overline{G(\mathcal{K})^\lambda}:= \bigsqcup_{\mu\in P^+, \mu \leq \lambda}G(\mathcal{K})^\mu$. Denote by $Gr:=G(\mathcal{K})/G(\mathcal{O})$ the affine Grassmannian and define $Gr^\lambda$ and $\overline{Gr^\lambda}$ similiarly.

Now let $\lambda,\mu,\nu\in P^+$ and $w\in W$ such that $\lambda:= w(\nu-\mu)\in P^+$. (Of course $W$ is the Weyl Group of $G$).

Then one has the multiplication map $$m:\overline{G(\mathcal{K})^\lambda}\times_{G(\mathcal{O})} Gr^\mu\to \overline{Gr^{\lambda+\mu}}$$

Now my question:

Is it true that $$m([g,x])\in Gr^\nu\Rightarrow g\in G(\mathcal{K})^\lambda$$ or even more $$m^{-1}(Gr^\nu)=[G(\mathcal{O})\lambda, [w\mu]] $$

I can prove both statements in some cases with simple calculations but i wonder if there is a "nice" proof in the general case. I would appreciate any hint or reference.