Some questions about specific topological embeddings

Question

It often happens in algebraic topology that we have a construction on, say, a space $X$, usually obtained by taking a product and then a quotient. Then we may have a "natural" injection of $X$ in the constructed space. My question is: under these circumstances, is there a general method to prove that this injection is indeed a (topological) embedding?

Let me list some examples related to mapping cones and cylinders:

1. Consider the unreduced cone $C'X:=X\times I/X\times1$. Then the canonical injection $X\hookrightarrow C'X, x\mapsto[x,0]$ is an embedding. This I can prove directly.
2. More generally, consider the reduced mapping cone on a pointed space $(X,*)$ defined by $CX:=X\times I/(X\times1\cup*\times I)$. The canonical injection $X\hookrightarrow CX, x\mapsto[x,0]$ is also an embedding. I cannot seem to find a proof of this...
3. Another situation arises in cofibrations. Let $i:A\subseteq X$. The mapping cylinder is $M(i):=A\times I\cup_i X$ obtained by identifying $(a,0)\sim i(a)$. We have an injection $M(i)\hookrightarrow X\times I$ defined in the obvious way. Is this map an embedding?

Explicitly, my question is: (1) How do I prove the above 2 and 3? (2) Is there a general method to tackle these questions?

Answer 1

Question (1):

2.) Let $p : X \times I \to CX$ denote the quotient map. It identifies $X' = X \times \{ 1\} \cup \{ *\} \times I$ to a point.The canonical injection $i : X \to CX$ is continuous. To see that $i$ is an embedding, we consider a closed $A \subset X$ and show that $i(A)$ is closed in $i(X)$.

Case 1. $* \in A$. Then $B = X \times \{ 1 \} \cup A \times I$ is closed in $X \times I$. Since $X' \subset B$, we have $p^{-1}(p(B)) = B$. Hence $p(B)$ is closed in $CX$ because $p$ is a quotient map. But $p(B) \cap i(X) = i(A)$.

Case 2. $* \notin A$. Then $B = A \times [0,1/2] $ is closed in $X \times I$. Since $X' \cap B = \emptyset$, we have $p^{-1}(p(B)) = B$ whence $p(B)$ is closed. Again $p(B) \cap i(X) = i(A)$.

Note, however, that $i(X)$ is closed in $CX$ if and only if $\{ \ast \}$ is closed in $X$. In contrast, in the unreduced cone $i(X)$ is always closed.

3.) Let $j : M(i) \to X \times I$ denote the unique map such that $j([x]) = (x,0)$ for $x \in X$ and $j([a,t]) = (a,t)$ for $(a,t) \in A \times I$. It is always an injection whose image is $M' = j(M(i)) = X \times \{ 0 \} \cup A \times I$. Moreover, it is an embedding if (a) $A$ is closed or (b) $i : A \to X$ is a cofibration. (a) is trivial. The highly nontrivial proof of (b) can be found in

Strøm, Arne. "Note on cofibrations II." Mathematica Scandinavica 22.1 (1969): 130-142.

See Lemma 3 of this paper and the comment following the proof. Note that there are cofibrations $i : A \to X$ where $A$ is not closed, but only with non-Hausdorff $X$.

If neither $A$ is closed nor $i$ is a cofibration, we get counterexamples like $X = I, A = [0,1)$.

Question (2):

I do not see a general method.