Question about quotient group of quotient group, and composition of quotient mappings I am still inching my way through Seth Warner's "Modern Algebra" (1965), and I have reached exercise 12.17 (b).
Part 12.17 (a) is the following:

Let $R$ be an equivalence relation on $E$ compatible with a composition $\triangle$ on $E$, and let $S$ be an equivalence relation on $E/R$ compatible with the induced composition $\triangle_R$. Let $T$ be the relation on $E$ satisfying $x \mathrel T y$ if and only if $[x]_R \mathrel S [y]_S$. Prove that $T$ is an equivalence relation on $E$ compatible with $\triangle$, and that there is a unique isomorphism $g$ from $(E/R)/S$ onto $E/T$ satisfying $g \circ \varphi_S \circ \varphi_R = \varphi_T$.

In the above, $\varphi_X$ denotes the quotient mapping $\varphi_X (x) = [x]_X$ where $[x]_X$ is the equivalence class of $x$ under $X$.
This bit's simple. Please do not give a solution for this in the answer, as I don't need it. I've done this bit.
Now, 12.17 (b) is:

Derive the theorem of Example 12.14 from (a).

Right, so this is Example 12.14:

If $H$ is a normal subgroup of a group $G$, if $K$ is a normal subgroup of $G/H$, and if $L = \varphi^\gets_H(K)$, then $L$ is a normal subgroup of $G$, and there is an isomorphism $f$ from $(G/H)/K$ onto $G/L$ satisfying $f \circ \varphi_K \circ \varphi_H = \varphi_L$.

Straightforward to start with: $H$ and $K$ induce and are induced by congruence relations on $G$ (that is, equivalence relations compatible with the group operation of $G$). Thus we can use the given definitions of $H$ and $K$ to use them as the protagonists in the drama that is the theorem proved in 12.17 (a).
On comparing the structure of 12.17 (a) and example 12.14, we arrive at the fact that the congruence relation $T$ whose existence is demonstrated in 12.17 (a) does indeed correspond with a normal subgroup $L$ such that $f \circ \varphi_K \circ \varphi_H = \varphi_L$.
But what I am having trouble with is identifying $L$ with $\varphi^\gets_H(K)$.
We have that:
$$\varphi^\gets_H(K) = \{x \in G: \varphi_H (x) \in K\} = \{x \in G: x H \in K\}$$
I now believe that all I need to do is show that $\{x \in G: x H \in K\}$ is $[e]_T$, that is, the equivalence class of the identity element of $G$ by $T$.
This is not homework for a marked assignment, so (at this stage) there is no immediate incentive to apply this to a qualification. It is self-study, just me being too lazy to think it through for myself. I am refreshing and enhancing where I left off at the end of my maths degree some decades ago, and I would dearly like to be able to turn over the page in "Modern Algebra", as it's been open at this page for a good week now.
EDIT: This will make clearer what I am attempting to do.
Let $e$ be the identity element of $G$.
Let $R$ be the congruence relation defined by $H$ in $G$.
Let $S$ be the congruence relation defined by $K$ in $G / H$.
Let $T$ be the relation on $G$ defined as:
$\forall x, y \in G: x \mathrel T y \iff x H \mathrel S y H$
Now $T$ is known (or easily shown, I did this in a different exercise) to be a congruence relation on $G$.
Hence the equivalence class under $T$ of $e$, that is $[e]_T$, is a normal subgroup of $G$.
Let us call this normal subgroup $L$, that is: $L := [e]_T$.
By what I have proved in 12.17 (a), I have that there exists a unique isomorphism $\phi$ from $(G / H) / K$ to $G / L$ which satisfies $\phi \circ \varphi_K \circ \varphi_H = \varphi_L$ where $\varphi_K$, $\varphi_H$ and $\varphi_L$ denote the quotient epimorphisms.
But what I have not been able to do is show that the $L$ which I have created above is the same thing as $\varphi^\gets_H(K)$.
I have accepted the correction from Arturo Magidin about the nature of $\varphi^\gets_H(K)$, but I am no closer to showing that $\varphi^\gets_H(K) = [e]_T$.
 A: With the help of comments posted by Arturo Magidin, I have completed the proof as follows.
Let $e$ be the identity element of $G$.
Let $R$ be the congruence relation defined by $H$ in $G$.
Let $S$ be the congruence relation defined by $K$ in $G / H$.
Let $T$ be the relation on $G$ defined as:
$$\forall x, y \in G: x \mathrel T y \iff x H \mathrel S y H$$
It is known (and has been proved a priori) that $T$ is a congruence relation on $G$.
Hence as a congruence relation on a group induces a normal subgroup, the equivalence class under $T$ of $e$, that is $[e]_T$, is a normal subgroup of $G$.
Then we have:
$L = \varphi_H^{-1} [K]$ (by hypothesis)
$= \{x \in G: \varphi_H (x) \in K\}$ (by definition of preimage of subset under mapping)
$= \{x \in G: x H \in K\}$ (by definition of quotient mapping)
Recall that $H$ is the identity of $G / H$.
Then as $K$ is a subgroup of $G / H$, we have that $H \in K$, as the identity of a subgroup is the same as the identity of its parent group.
Then:
$x \in [e]_T$
$\leftrightsquigarrow x \mathrel T e$ (by definition of equivalence class)
$\leftrightsquigarrow x H \mathrel S e H$
$\leftrightsquigarrow x H \in K$ by definition of $S$, and a priori $H \in K$
That is:
$$[e]_T = L$$
and so $L$ is a normal subgroup of $G$.
We can identify:
$\varphi_R \equiv \varphi_H$ as $R$ is the congruence relation defined by $H$ in $G$
$\varphi_S \equiv \varphi_K$ as $S$ is the congruence relation defined by $K$ in $G / H$
$\varphi_T \equiv \varphi_L$ as a congruence relation on a group induces a normal subgroup
Now a priori we have that there exists a unique isomorphism $f$ from $(G / H) / K$ to $G / L$ which satisfies $f \circ \varphi_S \circ \varphi_R = \varphi_T$
Thus using the above identfications:
$$f \circ \varphi_K \circ \varphi_H = \varphi_L$$
$\blacksquare$
