Is inclusion in a cone an embedding? In Tom Dieck's algebraic topology, pg 92,  he defines and states 

The (pointed) cone $CX$ over $X$ is $X \times I/ X \times 0 \cup * \times I$. The inclusion $i_1^X: X \rightarrow CX , x \mapsto (x,1)$ is an embedding. 

Why is this true? In fact if we take $C$ to be closed in $X$, let $q:X \times I \rightarrow CX$ denote the quotient map, then $$q^{-1}[i_1[C]]= C \times 1 \cup * \times I$$ if $* \in C$. $i_1$ is not necessarily a closed map to its image, hence not an embedding. 
Note: no where does Tom Dieck states he is working in category of CGWH spaces. Otherwise this statement would be true as $*$ is closed in $X$. 
 A: Let $X' = q(X \times 1) \subset CX$. We have to show that $j : X \to X', j(x) = q(x,1)$, is a homeomorphism. It is obviously a continuous bijection. Let $A \subset X$ be closed. We want to show that $j(A)$ is closed in $X'$. This means that there exists $B \subset CX$ such that $B \cap X' = j(A) = q(A \times 1)$ and $q^{-1}(B)$ is closed in $X \times I$.
Case 1: $\ast \notin A$.
Take $B = q(A \times 1)$. Then $q^{-1}(B) = A \times 1$ is closed in $X \times I$ and $B \cap X' = q(A \times 1) \cap q(X \times 1) = q(A \times 1)$.
Case 2: $\ast \in A$.
Take $B = q(A \times I)$. Then $q^{-1}(B) = A \times I \cup X \times 0$ is closed in $X \times I$ and $B \cap X' = q(A \times I) \cap q(X \times 1) = q(A \times 1)$. Note that $q(A \times I) = q(A \times \{0, 1 \} \cup \ast \times I) \cup q((A \backslash \ast) \times (0,1)) = q(A \times 1) \cup  q((A \backslash \ast) \times (0,1))$.
BUT: $i_1^X : X \to CX$ is in general not a closed embedding. It is one if and only if $q^{-1}(X') = X \times 1 \cup \ast \times I$ is closed in $X \times I$. The latter is equivalent to $\ast$ being closed in $X$.
