Every space $X$ can be identified with the closed subspace of the reduced cone $\operatorname{C}(X)$ of $X$. For any pointed space $(X,x_0)$, we define the cone $(\operatorname{C}(X),*)$ of $X$ to be the smash product $(X\wedge I, *)$ where the base point of $I$ is assumed to be $0$. The map $\mathbf{i} : X\to \operatorname{C}(X)$ defined by $\mathbf{i}(x) = x\wedge 1$, $x\in X$, is a continuous bijection from $X$ to $\mathbf{i}(X)$. In fact, $\mathbf{i}= \pi_\wedge \circ (Id_X ,C_1)$,
$$X\mathop {‎\longrightarrow‎}\limits^{(Id_X ,C_1)} X\times I \mathop {‎\longrightarrow‎}\limits^{\pi_\wedge}X\wedge I $$
Is it true that $\mathbf{i}\vert_X$ is a homeomorphism and $\mathbf{i}(X)$ is closed in $X\wedge I$ ? 
 A: The bottom of the unreduced cone is closed since it is the preimage of $1$ under the continuous projection from the  unreduced cone to the interval. But now the union of $x_0 \times I$ and the bottom of the unreduced cone is the whole equivalence class of the bottom of the reduced cone but this is closed as the union of two closed sets. Therefore the bottom of the reduced cone is closed, since its preimage in the unreduced cone is closed.
The map you described is also an isomorphism to its image, because every open subset $U$ of $X$ gives an open subset of $CX$ by looking at the equivalence class of $U \times [1,0)$, which is an open subset of the cone. Therefore we can conclude that the inclusion is also an open map to its image, since the subspace topology of the bottom is given by taking the intersection of an open subset of the whole cone with the bottom.
A: $(\operatorname{Id}_X, c_1)$ is a homeomorphism between $X$ and $X \times \{1\}$, for any space, because it has a continuous inverse (projection restricted).
$\pi_{\land}|_{X \times \{1\}}$ is 1-1 (as it intersects the identified set $A = (\{x_0\} \times I) \cup (X \times \{0\}$ only in $(x_0,1)$) and continuous as the restriction of a continuous map. If $\pi_{\land}$ were a closed map (which is the case if $\{x_0\}$ is assumed to be closed, because then we identify a closed set to a point and such maps are closed quotient maps), it stays closed when we restrict to the closed set $X \times \{1\}$ (as $\{1\}$ is closed in $I$). And then we both have that $pi_{\land}|_{X \times \{1\}}$ is a homeomorphism (as a 1-1 closed continuous map) and the image is closed.
I think it fails for non-closed $\{x_0\}$, because if $p \in X, p \neq x_0$ satisfies $p \in \overline{\{x\}}$, we see that e.g. $(p,\frac{1}{2}) \in \overline{i[X]} \setminus i[X]$.
