Closed inclusion

Question

I have a simple question in the context of (co)fibrations in the context of Model Categories: Why on the page $52$ in the snippet below

$$g^{-1}(d)$$ must be a single point not in the image of $A$ ? The definition of a $T_1$ closed inclusion is given first.

Answer 1

It's because of the way the pushout square is constructed. $D$ must be the disjoint union of $B$ and $C$, with every point $a\in A\subset B$ identified with its image $f(a)\in C$, and $g$ and $j$ are the two inclusions composed with the quotient map for that identification. In other words, $D=\frac{B\sqcup C}{\sim}$, where ${}\sim{}\subset( B\sqcup C)\times (B\sqcup C)$ contains the pairs of the form

• $(a,a')$ where $a,a'\in A$ and $f(a)=f(a')$
• $(a,f(a))$ and $(f(a),a)$ for all $a\in A$
• $(x,x)$ for all $x\in B\sqcup C$,

and $g,j$ are the restriction of the quotient map $B\sqcup C\to D=\frac{B\sqcup C}{\sim}$.

Thus a point $d$ that is not in the image of $j$ is an equivalence class that does not contain an element of $C$, thus it must the equivalence class of a point in $B\setminus A$, and these equivalence classes are singletons.