TITLE: (Certain) colimit and product in category of topological spaces
QUESTION [6 upvotes]: Consider the diagram 
$$(*):\;\;\;X_0 \stackrel{i_0}\hookrightarrow X_1 \stackrel{i_1}\hookrightarrow  X_2 \stackrel{i_2}\hookrightarrow \cdots $$
in category of topological spaces. Denote $I$ the unit interval $[0,1]$ with the standard topology.
Then we also have
$$(*)\times I:\;\;\;X_0\times I \stackrel{i_0\times 1_I}\hookrightarrow X_1 \times I \stackrel{i_1\times 1_I}\hookrightarrow  X_2 \times I \hookrightarrow \cdots $$
and colimits of those two diagrams, which are basically
$$\mathrm{colim}(*)=\left(\coprod_{n =0}^\infty X_n\right)/{\sim}, \;\;\;  x \sim i_j(x), \; j \in \mathbb{N}, \; x \in X_j,$$
$$\mathrm{colim}((*)\times I)=\left(\coprod_{n =0}^\infty X_n \times I\right)/{\sim}, \;\;\; (x,t) \sim (i_j(x),t), \; j \in \mathbb{N}, \; x \in X_j, t \in I.$$
(Question 0: is this correct?)
I am interested in the map $$f:(\operatorname{colim}(*))\times I \rightarrow \mathrm{colim}((*)\times I)$$
defined as $([x],t)\mapsto [(x,t)]$, where $[-]$ denotes taking the equivalence classes with respect to the considered equivalences. Obviously it is correctly defined as a map.
Question 1: Is this map continous? That is, are the two objects homeomorphic? (I believe one can obtain the inverse map $f^{-1}$ using the universal property of $\mathrm{colim}((*)\times I)$, hence $f^{-1}$ is continuous).
I have more information about the maps $i_j$. For example, they are closed maps.
This is a part of a homework assignment, so I would appreciate hints instead of full answer.
(Maybe some context and motivation: My goal is to define a homotopy $F:\mathrm{colim}(*))\times I \rightarrow \mathrm{colim}(*)$ using easy-to-describe homotopies $F_n:X_n\times I \rightarrow X_{n+1}$, but these induce "only" a map $(\operatorname{colim }F_n): \operatorname{colim}((*)\times I) \rightarrow \operatorname{colim}(*)$, so I am trying to change the domain somehow - so ultimately, I need only the composite map $(\operatorname{colim }F_n) \circ f$ to be continuous.)

REPLY [3 votes]: There is a natural continuous map 
$\varphi:\text{colim}(X_n×I)\to\text{colim}(X_n)\times I$ whose existence is indicated by the following diagram
$$\begin{array}{ccc}
\coprod(X_n×I) & → & \text{colim}(X_n×I) \\
\downarrow & & \downarrow\\
\left(\coprod X_n\right)×I & \xrightarrow{q×1} & \text{colim}(X_n)×I
\end{array}$$
where the left vertical map is induced by $X_n×I→\left(\coprod X_n\right)×I$. It's a standard exercise to show that the left map is a homeomorphism, so it doesn't really matter if we regard some $(x,t)$ an element of the first or the second space. Then $φ$ is induced as a continuous map from the quotient because if two points $(x,t)$ and $(y,s)$ are identified via the generating relations, then $s=t$ and $y=i_j(x)$, thus $[x]=[y]$. This $φ$ sends $[(x,t)]$ to $([x],t)$ and it is easy to show that $φ$ is bijective. This gives us the inverse $\psi$ which send $([x],t)$ to $[(x,t)]$, but there doesn't seem to be anything which suggests continuity of $ψ$.
The trick is to show that the map $q×1$ is a quotient map, as this will make the continuous bijection $φ$ a homeomorphism. There are two proofs for this fact, one more direct and elementary proof, and another proof which is more categorical and involves more diagrams and less topology, but both use the fact that $I$ is locally compact at some point.