The $0$-Hecke algebra $\mathcal{H}_0(S_{n+1})$ is generated by $\{ h_1, \ldots,h_n\}$ satisfying
i) $h_i^2=-h_i$
ii) $h_ih_j=h_jh_i$ if $|j-i| > 1$
iii) $h_ih_{i+1}h_i=h_{i+1}h_ih_{i+1}$.
Now, suppose $R$ is an algebra with basis $\{e_w : w \in S_{n+1}\}$ and let $\mathcal{R}$ be the subalgebra of $\operatorname{Hom}(R,R)$ generated by $\theta_1,\ldots,\theta_n:R \to R$ given by
$$\theta_{i}(e_w) = \left \{ \begin{matrix} e_{s_iw} & \text{if } \ell(s_iw) > \ell(w), \\ \\ -e_w & \text{if } \ell(s_iw) < \ell(w). \end{matrix} \right. $$
If we show that the $\theta_i$'s also satisfy i), ii) and iii), then does it follow that $\mathcal{R} \cong \mathcal{H}$? The reason I ask is because after showing all of this, Mathas in his book Iwahori-Hecke Algebras and Schur Algebras of the Symmetric Group, says that there's only a subjective hom $\mathcal{H} \to \mathcal{R}$. So why aren't they isomorphic?