Join of closed embeddings is a closed embedding An exercise from James's book General Topology and Homotopy Theory asks the reader to prove that if $\phi_1:X_1 \to Y_1$ and $\phi_2:X_2 \to Y_2$ are closed topological embeddings, then $\phi_1 * \phi_2:X_1 * X_2 \to Y_1 * Y_2$ is a closed topological embedding.
He did not explicity describe $\phi_1 * \phi_2$, and this is my first time learning about it. I gather that it is the map on $X_1 * X_2$ induced by $\pi \circ [(\text{Id}_I \times \phi_1 \times \phi_2) \sqcup \phi_1 \sqcup \phi_2]: (I \times X_1 \times X_2) \sqcup X_1 \sqcup X_2 \to Y_1 * Y_2$, where $\pi: (I \times Y_1 \times Y_2) \sqcup Y_1 \sqcup Y_2 \to Y_1 * Y_2$ is the quotient map.
First I'm trying to show that $\phi_1 * \phi_2$ should be an embedding (the closedness does not seem too hard to get from here). My idea was to use the characteristic property of an embedding, letting $f: W \to Y_1 * Y_2$ and $\tilde{f}:W \to X_1 * X_2$ such that $f = (\phi_1 * \phi_2) \circ \tilde{f}$ as set maps, and show that $f$ is continuous iff $\tilde{f}$ is. It's troublesome that I'm looking at arbitrary maps into quotients and coproducts, though.
One thing I'm trying is to look at the map $F$ which is $f$ except without the quotient at the end: i.e. $F = [(\text{Id}_I \times \phi_1 \times \phi_2) \sqcup \phi_1 \sqcup \phi_2] \circ \tilde{f}$. $F$ is continuous iff $f$ is. 
Any tips?
 A: For convenience, we may assume that the maps $\phi_i$ are inclusions of closed subspaces $X_i\subseteq Y_i$. These maps induce a closed inclusion $\phi:X_1\times X_2\times I\hookrightarrow Y_1\times Y_2\times I$. Seeing $X_1*X_2$ as a quotient of $X_1\times X_2\times I$ via the quotient map $q_X$ identifying $(x_1,x_2,0)\sim(x_1,x'_2,0)$ as well as $(x_1,x_2,1)\sim(x'_1,x_2,1)$, and similarly for the quotient map $q_Y$, it is easy to show that the induced map $\phi_1*\phi_2$ such that $(\phi_1*\phi_2) q_X=q_Y\phi$ is injective. It is closed because given a closed and saturated set $C\subseteq X_1\times X_2\times I$, its saturation $\bar C$ is closed as well. This is evident from the way the saturation is built: If $C$ contains some point $(x_1,x_2,0)$, then it contains a product $\pi_1(C\cap X_1×X_2×\{0\})\times X_2\times\{0\}$, and then we add the set $\pi_1(C\cap X_1×X_2×\{0\})\times Y_2\times\{0\}$ which is closed in $Y_1\times Y_2\times I$.

Addendum: Closedness is essential here. The claim need not hold when the subspaces are not closed. Take for example $X_1=Y_1$ as the one-point-space, and $X_2=[0,1)\subset I=Y_2$. Then the joins are the cones $C[0,1)$ and $CI$. Now the image of $U=\{(x,s)\in X_2\times I\mid x+s<1\}$ is open in $C[0,1)$, but it's not the intersection of $(\phi_1*\phi_2)(C[0,1))$ with some open set in $CI$.
