How to prove a lemma of Rouquier on the dimension of triangulated categories?

Question

In the paper of Rouquier on the dimension of triangulated categories (found here) lemma 3.5 says:

Lemma Let $\mathcal{T}$ be a triangulated category and let $\mathcal{T}_1$ and $\mathcal{T}_2$ be triangulated subcategories of $\mathcal{T}$ such that $\mathcal{T}=\mathcal{T}_1\diamond\mathcal{T}_2$. Then $dim\mathcal{T}\leq dim\mathcal{T}_1+dim\mathcal{T}_2+1$.

There is no proof given in the text and it says that the lemma is "clear". However, at the moment, I really can't see how one can proceed. My guess is that by assuming that $\mathcal{T}_1=\langle M_1\rangle_{1+dim\mathcal{T}_1}$ and $\mathcal{T}_2=\langle M_2\rangle_{1+dim\mathcal{T}_2}$, one can find an object $M$ such that $\mathcal{T}=\langle M\rangle_{2+dim\mathcal{T}_1+dim\mathcal{T}_2}$.

Is there something trivial that I am missing? Any help would be very much appreciated.

Answer 1

If $\mathcal{T}_{1}=\langle M_{1}\rangle_{d_{1}+1}$ and $\mathcal{T}_{2}=\langle M_{2}\rangle_{d_{2}+1}$, then $\mathcal{T}_{1}\ast\mathcal{T}_{2}\subseteq\langle M_{1}\oplus M_{2}\rangle_{d_{1}+d_{2}+2}$.

Sketch proof: Since $\mathcal{T}_{i}\subseteq\langle M_{1}\oplus M_{2}\rangle_{d_{i}+1}$, it suffices to show that $$\langle M_{1}\oplus M_{2}\rangle_{s}\ast\langle M_{1}\oplus M_{2}\rangle_{t}= \langle M_{1}\oplus M_{2}\rangle_{s+t}.$$

This follows by induction on $s$ and $t$, using the fact that $\ast$ is associative (i.e., $\mathcal{A}\ast(\mathcal{B}\ast\mathcal{C})=(\mathcal{A}\ast\mathcal{B})\ast \mathcal{C}$ for triangulated subcategories $\mathcal{A},\mathcal{B},\mathcal{C}$ of $\mathcal{T}$), which follows from the octahedral axiom.