Showing that the natural map $X_0\cup_f X_1\to X_0\cup_F (X_1\times I)$ is a topological embedding Let $X_0,X_1$ be topological spaces, $A\subset X_1$, and $f:A\to X_0$, $F:A\times I\to X_0$ continuous maps with $F(a,0)=f(a)$. Consider two adjunction spaces $X_0\cup_f X_1$ and $X_0\cup_F (X_1\times I)$. By the condition $F(a,0)=f(a)$, we have a well-defined continuous injection  $X_0\cup_f X_1\to X_0\cup_F (X_1\times I)$. Is this map an embedding, assuming that $(X_1,A)$ has the homotopy extension property?
This is implicitly used in Hatcher's book, Chapter  0, assuming that $(X_1,A)$ has the homotopy extension property (Proposition 0.18 and Exercise 26), but I can't see why the map $X_0\cup_f X_1 \to X_0\cup_F (X_1\times I)$ is an embedding.
 A: This is indeed implicit in the discussion. The "retraction" of $X_0\cup_F(X_1\times I)$ onto $X_0\cup_fX_1$ witnesses that this is in fact an embedded subspace. "Retraction" is in quotation marks, because that notion only makes sense a posteriori, once we know this is a subspace. So let me spell out what I mean:
There is a retraction $r\colon X_1\times I\rightarrow A\times I\cup X_1\times\{0\}$ (this is equivalent to $(X_1,A)$ having the homotopy extension property). In this case, a map on $A\times I\cup X_1\times\{0\}$ is continuous if it is continuous when restricted to $A\times I$ and $X_1\times\{0\}$ each (this is clear if $A\subseteq X_1$ is closed; the general case is the rather technical argument which Hatcher has transferred to the appendix). There is a continuous map $\tilde{F}\colon A\times I\cup X_1\times\{0\}\rightarrow X_0\cup_fX_1$, given by $\tilde{F}(a,t)=[F(a,t)]$ for $(a,t)\in A\times I$ and $\tilde{F}(x_1,0)=[x_1]$ for $(x_1,0)\in X_1\times\{0\}$. This is well-defined on the intersection since $[F(a,0)]=[f(a)]=[a]$ for all $a\in A$ by the definition of $X_0\cup_fX_1$. By the preceeding remark, $\tilde{F}$ is continuous.
The continuous composition $\tilde{F}r\colon X\times I\rightarrow X_0\cup_fX_1$ and the canonical embedding $X_0\rightarrow X_0\cup_fX_1$ are compatible in the sense that $\tilde{F}r(a,t)=\tilde{F}(a,t)=[F(a,t)]$ for $(a,t)\in A\times I$, so they glue together to a continuous map $R\colon X_0\cup_F(X_1\times I)\rightarrow X_0\cup_fX_1$. I leave it to you to check that the restriction of $R$ to the image of the canonical map $X_0\cup_fX_1\rightarrow X_0\cup_F(X_1\times I)$ is an inverse to it (when co-restricted to the image), proving that it is a homeomorphism onto its image, i.e. an embedding.
Loosely speaking, the point is that a retraction $X_1\times I\rightarrow A\times I\cup X_1\times\{0\}$ induces a retraction $X_0\cup_F(X_1\times I)\rightarrow X_0\cup_F(A\times I\cup X_1\times\{0\})$ and this latter subspace can be identified with $X_0\cup_fX_1$ since $X_1\times\{0\}$ gets attached via $F\vert_{X_1\times\{0\}}$, which corresponds to attaching $X_1$ via $f$, and attaching $A\times I\setminus\{0\}$ is merely "excess" as this gets identified to points in $X_0$ anyhow.
If we identify $X_0\cup_fX_1$ as subspace of $X_0\cup_F(X_1\times I)$ in this way, the map $R$ becomes a bona fide retraction. To this into 0.18, if you in fact have a deformation retraction of $X_1\times I$ onto $A\times I\cup X_1\times\{0\}$, a similar construciton induces a deformation retraction of $X_0\cup_F(X_1\times I)\rightarrow X_0\cup_fX_1$.
