How do I see that the map $X^n \to X^{n+1}$ of CW-building blocks is an embedding? In his book “A concise course in algebraic topology”, May defines a CW complex inductively as being the union of increasing subspaces $X^n$, where

*

*$X^0$ is discrete

*$X^{n+1}$ is the simultaneous pushout of attaching maps $X^n \stackrel{j_\alpha}{\leftarrow} S^{n}\hookrightarrow D^{n+1}$.

I do realize that $X^{n+1}$ is the quotient of the disjoint union $X^n \sqcup\left(\bigsqcup_\alpha D^{n+1}\right)$ subject to the relations $(\alpha, s)\sim j_\alpha(s)$ for any $s\in S^n$, but I fail to see why the resulting map $\eta\colon X^n \to X^{n+1}$ into the pushout should be an embedding.
Question
How do we know, at least in $\mathbf{Top}$, that $\eta$ in the pushout diagram below is an embedding if we know that each $\iota_\alpha$ is an embedding? (I've added indices to clarify which objects occur repeatedly in the colimit diagram)
$$\require{AMScd}
\begin{CD}
S^n_\alpha @>{\iota_\alpha}>> D^{n+1}_\alpha \\
@VV{j_\alpha}V @VV{\eta_\alpha}V\\
X^n @>{\eta}>> X^{n+1} 
\end{CD}
$$
I've tried to prove this for regular monomorphisms in general categories, but have not produced something promising in the past hour. Perhaps I'm overseeing something obvious.
 A: In fact we have a pushout diagram
$\require{AMScd}$
\begin{CD}
X^n @>{\iota'}>> X^{n+1} \\
@A{j}AA @A{J}AA\\
\bigsqcup_\alpha S^{n} @>{\iota}>> \bigsqcup_\alpha D^{n+1}\end{CD}
Unfortunately there is no way to prove that $\iota'$ is an embedding by some general category theory. It is a property of $\mathbf{Top}$ that pushouts of embeddings are embeddings. Thus, since $\iota$ is an embedding, also $\iota'$ is one.
I shall not prove this general pushout theorem here, concerning pushouts and adjunction spaces in $\mathbf{Top}$ see for example tom Dieck's "General Topology", especially Proposition (1.8.1). However, the following is very easy to prove and covers your question:
Let $A \subset Y$ be a closed subspace with inclusion $i : A \to Y$. Let
$\require{AMScd}$
\begin{CD}
X @>{I}>> Z \\
@A{f}AA @A{F}AA\\
A @>{i}>> Y\end{CD}
be the pushout with $f : A \to X$. Then $I$ is a closed embedding.
The standard construction in $\mathbf{Top}$ gives $Z = X \cup_f Y$ which is the well-known adjunction space. Let $p : X \sqcup Y \to  X \cup_f Y$ be the quotient map identifying $a = i(a) \in Y$ with $f(a) \in X$, for all $a \in A$. We have $I  = p \mid_X$ and $F = p \mid_Y$. It is easy to verify that for $x \in X$
$$p^{-1}(p(x)) =  \{x\} \sqcup f^{-1}(x) .$$
This show that $I$ is injective because $I^{-1}(I(x)) = p^{-1}(p(x)) \cap X = \{x\}$.
Now let $C \subset X$ be closed. Then
$$p^{-1}(I(C)) = p^{-1}(\bigcup_{x \in C}I(x)) = p^{-1}(\bigcup_{x \in C}p(x)) = \bigcup_{x \in C}p^{-1}(p(x)) = \bigcup_{x \in C}(\{x\} \sqcup f^{-1}(x)) \\ = \bigcup_{x \in C}\{x\} \sqcup \bigcup_{x \in C}f^{-1}(x) = C \sqcup f^{-1}(\bigcup_{x \in C}\{x\}) = C \sqcup f^{-1}(C)$$
which is closed in $X \sqcup Y$ (recall that $A$ is closed in $Y$). Thus $I(C)$ is closed in $X \cup_f Y$.
