\cdot \frac{n-1}{2} \cdot \prod_{j=0}^{h-3}\left(\frac{n+1}{2^{j+2}}-1\right)
                \nonumber \\ 
            & = 
                \prod_{j=0}^{h-1}\left(\frac{n+1}{2^j}-1\right). 
        \end{aligned}$$ Where in the second equality we used that $\frac{n-3}{4}\ge \frac{2^{h-1}-1-3}{4}=2^{h-3}-1$.

We now show that $Tr_h$ is globally forcing for $h\geq 4$. To see this, suppose that $N^*_T(h) \leq (2^{-\binom {h}{2}} + o(1)) n^{h}$. From this gives $N^*_T(3) \le (1+o(1)) n^3 / 8$. As $h\geq 4$, the function $x^{h-2}$ is strictly convex and by the application of Jensen in we find $|N(e)| = (1 \pm o(1))\frac{n}{4}$ for almost all $e \in E(T)$. However, by a result of Chung and Graham (see property $P_4$ and Theorem 1 in [@CG]) this property implies that $T$ is quasirandom. Thus $Tr_h$ is globally forcing for $h\geq 4$.

To complete the proof of the proposition, it remains to show that any non-transitive $h$-vertex tournament $H$ with $h\geq 7$ is not globally forcing. To see this, let $V(H) = \{u_1,\ldots ,u_h\}$. For each $n\in {\mathbb N}$ construct an $n$-vertex tournament ${T_1}$ as follows. Let $V(T_1) = V = \{v_1,\ldots , v_n\}$ denote a set of order $n$ and let $V = \cup _{i\in [h]} V_i$ denote a partition with $|V_i| = \lfloor n/h \rfloor $ or $\lceil n/h \rceil $ for all $i\in [h]$. For each edge 