Notations
Let $L$ be a Lie algebra, $\mathfrak{T}(L)=\oplus_{i=0}^{\infty} T^i L$ the tensor algebra on $L$, and $\mathfrak{U}(L) = \mathfrak{T}(L)/J$ a universal enveloping algebra of $L$ where $J$ is the two sided ideal in $\mathfrak{T}(L)$ generated by all $x \otimes y - y \otimes x - [x, y]$ ($x, y \in L$). Denote the canonical homomorphism $\mathfrak{T}(L) \rightarrow \mathfrak{U}(L)$ by $\pi$. For brevity, write $\mathfrak{T} = \mathfrak{T}(L)$, $\mathfrak{U} = \mathfrak{U}(L)$, and $T^m = T^m(L)=\underbrace{L \otimes L \otimes \cdots \otimes L}_{m \text{ times}}$.
Question and My Thoughts
I'm reading on Introduction to Lie algebras and Representation Theory (3rd edition, p.91) written by J.Humphreys. In the preliminary stage of the formulating the PBW theorem, he define a filtration on $\mathfrak{T}$ by $T_m = T^0 \oplus T^1 \oplus \cdots \oplus T^m $ and set $U_m = \pi(T_m)$, with the following claim.
Since $\pi$ maps $T^m$ into $U_m$, the composite linear map $\phi_m \colon T^m \rightarrow U_m \rightarrow U_m / U_{m-1}$ makes sense. It is surjective, because $\pi(T_m - T_{m-1}) = U_m - U_{m-1}$.
However, I think $\pi(T_m - T_{m-1}) \neq U_m - U_{m-1}$ in general. For example, let $x$ and $y$ be linearly independent elements of $L$. Then $x \otimes y - y \otimes x \in T_2 - T_1$, but $\pi(x \otimes y - y \otimes x) = \pi([x, y]) \in U_1 = \pi(T_1)$. Of course, it is straightforward to show $\pi(T_m - T_{m-1}) \supset U_m - U_{m-1}$ so the surjectivity of $\phi_m$ follows.
Is there any fault in my reasoning?