To avoid confusion, let us give the following definitions:

**Definitions**:

- Let $ \Lambda $ be $ \mathbb{R}, \mathbb{C} $ or the quaternions $ \mathfrak{Q} $ and let $G$ be a topological group. A $\Lambda G $ -space is a finite-dimensional vector space $V$ together with a morphism of topological groups $ \theta \ \colon G \longrightarrow Aut(V) $. Such a $V$ is also called a representation of $G$ over $\Lambda$.
- Let $V,W$ be $\Lambda G $ -spaces. A $\Lambda G $ -morphism $ f \ \colon V \longrightarrow W $ is a $G$-morphism (i.e. $f$ is compatible with the action of $G$) which is a $ \Lambda $-linear map.

**Fact**: If $ \theta \ \colon G \longrightarrow Hom_{\Lambda}(V,V) $ is a representation, then the image of \[ I = \int_{G} \theta = \int_{g \in G} \theta(g) \in Hom_{\Lambda}(V,V) \]is $V^{G}$, the subspace of $G$-invariant elements of $V$.

**Assertion**: (Using the above fact, show that) If $G$ is a compact topological group and $V$ is a $ \Lambda G$-space, then $V$ is a projective module (i.e. for every $ \Lambda G $-map $ \alpha \ \colon V \longrightarrow Y $ and every surjective $ \Lambda G $-map $ \beta \ \colon X \longrightarrow Y $ there exists a $ \Lambda G $-map $ \gamma \ \colon V \longrightarrow X $ such that $ \alpha = \beta \circ \gamma$).