TITLE: Is the Brauer correspondence injective ? 
QUESTION [8 upvotes]: Let $G$ be a finite group and $k$ a field. In modular representation theory the Brauer correspondence establishes a bijection between the isomorphism classes of indecomposable $kG$-modules with trivial source and vertex $P$ and the isomorphism classes of non-zero projective indecomposable $k[N_G(P)/P]$-modules. This bijection is induced by the Brauer map 
$$M \mapsto \operatorname{Br}_P(M) := \frac{M^P}{\sum_Q tr_{Q}^P (M^Q)}$$
$(Q < P)$ where $M^P$ are the $P$-invariants and $tr^P_Q: M^Q \to M^P,\;\; m \mapsto \sum_{g \in P/Q}gm$ is the trace map. This was proved in Theorem 3.2 of the paper 
M. Broué: On Scott Modules and p-Permutation Modules: An Approach through the Brauer Morphism. Proc. Amer. Math. Soc. 93(1985), 401-408
and the statement can also be found in this paper of M. Wildon (end of p. 12). 
However, to my understanding, Broué's proof doesn't show that if $M_1,M_2$ aren't ismorphic then $\operatorname{Br}_P(M_1), \operatorname{Br}_P(M_2)$ aren't isomorphic as well (as $k[N_G(P)/P]$-modules). 
Question: Does someone know, if the Brauer correspondence $[M] \mapsto [\operatorname{Br}_P(M)]$ ($[-]$ denotes isomorphism classes) is injective ? If so, a proof or a reference which contains a proof is also welcome.

Let me just explain why I think  Broué's proof doesn't show injectivity in full strenght: He decomposes $kG \otimes_{kP}k = \bigoplus_{i=1}^n M_i$ into indecomposable $kG$-Modules whereby the ones with vertex $P$ are exactly those such that $\operatorname{Br}_P(M_i) \neq 0$. Assume this holds for $i=1,...m$. Then 
$$k[N_G(P)/P] = \operatorname{Br}_P(kG \otimes_Pk)= \bigoplus_{i=1}^m \operatorname{Br}_P(M_i)$$
and the $\operatorname{Br}_P(M_i),\;(i=1,...m)$ can be shown to be indecomposable. Since every indecomposable projective $k[N_G(P)/P]$-module is isomorphic to a direct summand of $k[N_G(P)/P]$, the Brauer correspondence is surely surjective. 
However, to my understanding, it may happen that $M_i,M_j$ aren't isomorphic, but $\operatorname{Br}_P(M_i) \cong \operatorname{Br}_P(M_j)$ (i.e. this summand has multiplicity $>1$ in $k[N_G(P)/P])$.

REPLY [4 votes]: As Geoff Robinson's answer makes clear, the Brauer map is injective when restricted to indecomposable $p$-permutation $kG$-modules with vertex $P$, because it agrees with the Green correspondence. This is stated in Theorem 3.4 of the paper by Broué cited in the question. Natalie's answer mentions Thévenaz's book, which also has a proof of this fact.
If we allow non $p$-permutation modules (but still require fixed vertex $P$) then the map is not injective. For example, take $G = P = \langle g \rangle \cong C_8$ and let $k = \mathbf{F}_2$. Let $V_k$ be the unique $k$-dimensional indecomposable representation of $G$, for $1 \le k \le 8$. Then $V_3$ has a basis $v_1,v_2,v_3$ such that $v_i(g+1) = v_{i+1}$ for $i=1,2$ and $v_3(g+1) = 0$. Clearly $V_3^P = v_3$. Since 
$$v_2g^2 = v_2 + v_2(1+g^2) = v_2 + v_2(1+g)^2 = v_2$$ 
and $\mathrm{Tr}_{\langle g^2 \rangle}^P v_2 = v_2(1+g) = v_3$, we have $\mathrm{Br}_P(V_3) = 0$. A similar argument shows that $\mathrm{Br}_P(V_5) = \mathrm{Br}_P(V_7) = 0$. But since $V_3$, $V_5$ and $V_7$ have odd dimension, they all have full vertex $P$.
There is a related example in Example 1.3.6 of the note of mine linked in the question.