Suppose $X\cong Y$ and $A\cong B$. When is $X/A\cong Y/B$?
This is from one of my previous questions which was answered here.
The statement is true and it relies on the following lemma
Lemma. Let $f\colon X\to Y$ and $g\colon A\to B$ such that $f\circ i_A = i_B\circ g$. Then there exists unique $h\colon X/A\to X/B$ such that $h\circ p_A = p_B \circ f$, i.e. there is following commutative diagram $$\require{AMScd} \begin{CD} 0 @>>> A @>i_A>> X @>p_A>> X/A @>>> 0\\ & @VVgV @VVfV @VV\exists!\, hV \\ 0 @>>> B @>i_B>> Y @>p_B>> Y/B @>>> 0 \end{CD}$$
The function $h$ is defined by $h([x]_{X/A}) = [f(x)]_{Y/B}$, can this be deduced from the 5-lemma or we need to do a diagram chase type of proof.
By the commutativity of the left square, we can see that the map $p_B\circ f$ has kernel containing $A$, so factors uniquely through the quotient map $p_A:X\to X/A$.