Let $X\cup_f Y$ be an adjunction space and $q:X\coprod Y\rightarrow X\cup_f Y$ be the associated quotient map. Let $A\subseteq Y$ be closed and $f:A\rightarrow X$ be a continuous map (this is called the attaching map).
Note that there exists a canonical injection $\sigma$ from $X$ to the disjoint union
Note, $X \cup_f Y$ is the quotient space obtained by the relation $a\sim f(a)$ for all $a\in A$.
Problem:
The restriction of $q$ to $X$ is a topological embedding, whose image set $q(X)$ is a closed subset of $X\cup_f Y$.
My attempt: Clearly, $q|_X$ is continuous. I was able to show injectivity in a different post. So, in this post, I will that $q|_X$ is a closed map.
Let $B$ be a closed subset of $X\coprod Y$. Then I must show that
$q|_X(B)=q(B)$ is closed in the quotient space, equivalently, $q^{-1}(q(B))$ is closed in $X\coprod Y$.
I was able to show that
$\sigma_0^{-1}(q^{-1}(q(B)) = \sigma_0^{-1}(B)$ and that
$\sigma_1^{-1}(q^{-1}(q(B))=f^{-1}(\sigma_0^{-1}(B))$
and so $q^{-1}(q(B))$ is closed in the disjoint union.
Since $X$ is closed in the disjoint union, it follows that $q(X)$ is closed in the quotient space.
Is this correct? (please answer this)
Another question I have is, the continuous map, $f:A\rightarrow X$ (forgetting the identification), is still the map, $f:A\rightarrow X$, right?
Best Answer
OK, if $x \in X$ there is at most one $a \in A \subseteq Y$ such that $f(a)=x$ so that $a \sim x$ and $q(a)=q(x)$. No further identifications are made that concern $x$. That's why $q\restriction_X$ is injective (In my courses $f$ always went from $A$ (subset of $X$) to $Y$, which makes this confusing for me).
Indeed to see $q'=q\restriction_X: X \to X\cup_f Y$ is closed, we check that $q'[X]=q[X]$ is closed in $X\cup_f Y$ and this is equivalent to $q^{-1}[q[X]] \subseteq X \coprod Y$ being closed and by the final topology definition this in turn is equivalent to $\sigma_X^{-1}[q^{-1}[q[X]]]] \subseteq X$ being closed (but this is just $X$ so that's OK) and $\sigma_Y^{-1}[q^{-1}[q[X]]]$ being closed in $Y$, but this set equals $A$ which is closed in $Y$ (you don't mention it, but this is usually implicit if you say "adjunction space". So $q'[X]$ is closed in $X \cup_f Y$.
For closed $B$ of $X$ we see that $\sigma_X^{-1}[q^{-1}[q'[B]]] = B$, so closed in $X$ and $\sigma_Y^{-1}[q^{-1}[q'[B]]] = f^{-1}[B]$ which is closed in $A$ so closed in $Y$ under the (as said common) assumption that $A$ is closed in $Y$. In total one concludes that $q$ is a closed embedding of $X$ into $X \cup_f Y$.
So your approach is good, but again be careful in your identifications. One $X$ is not the same as the other, just equivalent under the $\sigma_X$.
It's quite similar to check that $q$ on $Y \setminus A$ also is an embedding, but an open one, if $A$ is closed in $Y$.
And finally, on $q[A]$ we can also define a version of $f$: $$\tilde{f}: q[A] \to X: \tilde{f}(q(a))=f(a)$$ which is also well-defined and continuous. I'm not sure if that is what you mean by the last sentence? Or do you mean $f': \sigma_Y[A] \subseteq X \coprod Y \to X$ defined by $f'(\sigma_Y(a))=f(a)$, which is also possible?