I understand the idea of a quotient algebra $A / I$ where $A$ is a $K$-algebra and $I$ is a two-sided ideal, i.e. I understand the projection map as an algebra morphism.
However, I'm unsure about how the relations defining generators for an ideal work.
Specifically, I'm looking at the universal enveloping algebra $U(L)$ of a Lie algebra $L$. I understand I want to quotient the tensor algebra $T(V)$ by the ideal generated by elements of the form
$\begin{equation*} XY - YX - [X,Y] \end{equation*}$
where $X, Y \in L$.
But why does it mean the same thing to say that the elements must fulfill the relation $XY - YX = [X,Y]$?
From what I know of group theory and cosets, I have a vague intuition about the ideal itself being our zero element in $T(L)/I$.
In short, why do elements of form $XY - YX - [X,Y]$ fulfill $XY - YX = [X,Y]$?
Let $L$ be a Lie algebra and $I\subset T(L)$ be the smallest ideal containing $\{xy-yx-[x,y]|x,y\in L\}$. Then by definition, $U(L)=T(L)/I$. Now every element of $L$ can be seen as an element of $U(L)$ via the map $$f:L\to U(L)$$ defined as the composite of the inclusion $i:L\to T(L)$ and the projection $\pi:T(L)\to U(L)$.
The claim is that $f(x)f(y)-f(y)f(x)=f([x,y])$ in $U(L)$ for every $x,y\in L$. This is the case because $xy-yx-[x,y]\in I$ by definition of $I$ and $$0=\pi(xy-yx-[x,y])=f(x)f(y)-f(y)f(x)-f([x,y]).$$