# (Certain) colimit and product in category of topological spaces

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.)

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Regarding question 1: there is a natural continuous map $\mathrm{colim} ((*) \times I) \to (\mathrm{colim} (*)) \times I$, and it is indeed the inverse of $f$. So $f$ is a homeomorphism as soon as it is continuous. – Zhen Lin Jun 13 '14 at 16:41
Homotopies are usually represented by $\pi_n$, at least where I've seen them. $H_n$ is more reserved for homology. – Sanath K. Devalapurkar Jun 13 '14 at 18:47

I asked a very similar question some time ago. The key fact is the following:

Let $X,Y$ be a topological spaces. Suppose $X$ comes with an equivalence relation $\mathcal{R}\subset X\times X$. Then the canonical map $$\frac{X\times Y}{\mathcal R\times \mathsf{1}}\longrightarrow \frac{X}{\mathcal{R}}\times Y,$$ is a continuous bijection (that may fail to be a homeomorphism). However, if $Y$ is locally compact Hausdorff, then it is a homeomorphism.

Here $\mathcal R\times \mathsf{1}$ designates the equivalence relation on $X\times Y$ whereby $(x,y)\sim (x',y')$ iff $x\mathcal{R}x'$ and $y=y'$. The proof of continuity is easy, it follows immediately form the universal properties of quotient spaces and products. To finish off your question, let $X=\coprod_nX_n$, $Y=I$ and $\mathcal R$ the equivalence relation defining the colimit of your first diagram. Then the colimit of your second diagram is the quotient space of $X\times I$ (because $\left(\coprod_nX_n\right)\times I$ is canonically isomorphic to $\coprod_n\left(X_n\times I\right)$) by the equivalence relation $\mathcal R\times\mathsf{1}$. The facts from the highlighted paragraph then tell us that the comparison map you defined is a homeomorphism.

This can be extended. The key fact is that the functor $$-\times I:\mathbf{Top}\to\mathbf{Top}$$ has a right adjoint $$\mathbf{Hom}(I,-):\mathbf{Top}\to\mathbf{Top}$$ where $\mathbf{Hom}(I,-)$ sends a space $X$ to the set of all continuous maps $I\to X$ equipped with the compact open topology. The unit interval is core compact (being locally compact Hausdorff). Having a right adjoint, the functor $-\times I$ commutes to all colimits, that is there are isomorphisms in Top (in other words, a homeomorphism) $$\mathrm{colim}_{\,\mathcal{I}}(F)\times I\to\mathrm{colim}_{\,\mathcal{I}}((-\times I)\circ F)$$ for any small diagram $F:\mathcal I\to\mathbf{Top}$.

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Very nice, as well as your related more detailed question. Can you give a hint how would one go about proving the part "if $Y$ is locally compact Hausdorff, then it is a homeomorphism"? Or possibly a reference? – Pavel Čoupek Jun 13 '14 at 19:35
@PavelC You can give a proof directly, it is not very difficult, but I have to admit I had a hard time proving it the first time around. The proof is written down in Switzer's Algebraic topology. – Olivier Bégassat Jun 13 '14 at 19:43

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.

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@PavelC: Oops, I didn't notice that Olivier has posted an answer while I was writing mine. His post answers your question. The only detail I could add is that the Hausdorffness of $I$ is not needed since local compactness is enough. – Stefan Hamcke Jun 13 '14 at 20:03