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Given diagrams of topological spaces $$X_0\rightarrow X_1\rightarrow\ldots$$ $$Y_0\rightarrow Y_1\rightarrow\ldots$$ $$Z_0\rightarrow Z_1\rightarrow\ldots$$ and maps $X_i\rightarrow Z_i$, $Y_i\rightarrow Z_i$, such that everything commutes. Taking the colimit yields a canonical map $$colim(X_i\times_{Z_i}Y_i)\rightarrow colim(X_i)\times_{colim(Z_i)}colim(Y_i)$$ from the colimit of the pullbacks to the pullback of the colimits. This map is a continuous bijection, since filtered colimits commute with finite limits in the category of sets and the forgetful functor from spaces to sets preserves all limits and colimits.

This answer of @StefanHamcke serves as a counterexample where the map is not a homeomorphism even in the case where the first three diagrams consist of closed inclusions.

I'm interested in mild point set topological restrictions on the spaces and maps, such that the canonical map is a homeomorphism. The answer linked suggests, that separation conditions will be necessary, so what e.g. about closed inclusions of T1 spaces, Hausdorff spaces, etc.?

Side question: Is this true in any convenient category of topological spaces like CGWH?

  • $\begingroup$ The counterexample you quote is already Hausdorff, if I am not mistaken. $\endgroup$
    – user43326
    Aug 26, 2015 at 6:15

2 Answers 2


Fiber products do commute with sequential colimits of closed embedding in CGWH, but one must remember that neither limits nor colimits in CGWH are the same as the corresponding limits and colimit in the category TOP of topological spaces. If you want to keep the fiber product and sequential colimits of TOP, here is a setting where this whould work:

Claim: Fiber products commute with sequential colimits of closed embeddings if the $X_i$'s and $Y_i$'s are Hausdorff locally compact and the $Z_i$'s are Hausdorff.

Proof: As mentioned in the question we have a continuous bijection $$ f:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i) $$ Our first step is to reduce to the case where $Z_i = \ast$ for every $i$. To do so consider the continuous bijection $$ g:colim_i(X_i \times Y_i) \longrightarrow colim_i(X_i) \times colim_i(Y_i) $$ By the universal property we may identify $colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i)$ with a subspace of $colim_i(X_i) \times colim_i(Y_i)$ (equipped with the subspace topology). On the other hand, since each $Z_i$ is Hausdorff the maps $h_i:X_i \times_{Z_i} Y_i \longrightarrow X_i \times Y_i$ are closed embeddings. We claim that the induced map $h:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i \times Y_i)$ is a closed embedding as well. First since sequential colimits of sets preserve injective maps it follows that $h$ is injective. Now let $A \subseteq colim_i(X_i \times_{Z_i} Y_i)$ be a closed set. Then the intersection of $h(A)$ with the $X_i \times Y_i$ coincides with $h_i(A \cap (X_i \times_{Z_i} Y_i))$, which is closed. Hence $h(A)$ is closed as desired. We may now conclude that the topology on $colim_i(X_i \times_{Z_i} Y_i)$ is the one induced from $colim_i(X_i \times Y_i)$. It will hence suffice to proce that the map $g$ is a homeomorphism. In other words, we may assume that $Z_i = \ast$.

Let $W$ be an open subset of $colim_i(X_i \times Y_i)$. We wish to show that $g(W)$ is open. For each $i$ let us denote by $W_i = W \cap (X_i \times Y_i)$, considered as an open subset of $X_i \times Y_i$. Let $(x,y) \in colim_i(X_i) \times colim_i(Y_i)$ be a point which belongs to $g(W)$. Then there exists an $i$ such that $x \in X_i$ and $y \in Y_i$ and we may find open neighborhood $x \in U_i \subseteq X_i$ and $y \in V_i \subseteq Y_i$ such that $U_i \times V_i \subseteq W_i$. Furthermore, since $X_i$ and $Y_i$ are locally compact we may assume in addition that the respective closures $\overline{U}_i$ and $\overline{V}_i$ are compact and that $\overline{U}_i \times \overline{V}_i \subseteq W$. Our next goal is to find open subsets $U_{i+1} \subseteq X_{i+1}$ and $V_{i+1} \subseteq Y_{i+1}$ such that $U_{i+1} \cap X_i = U_i$, $V_{i+1} \cap Y_i = V_i$, the closures $\overline{U}_{i+1}, \overline{V}_{i+1}$ are both compact, and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W$. To construct $U_{i+1}$ and $V_{i+1}$ we begin by choosing for every $(u,v) \in \overline{U}_i \times \overline{V}_i$ open neighborhoods $u \in A_{u,v} \subseteq X_{i+1}$ and $v \in B_{u,v} \subseteq Y_{i+1}$ such that $\overline{A}_{u,v}$ and $\overline{B}_{u,v}$ are compact and $\overline{A}_{u,v} \times \overline{B}_{u,v} \subseteq W_{i+1}$. For each $u \in \overline{U}_i$ we may then find a finite collection $B_{u,v_1}, ...,B_{u,v_n}$ which covers $\overline{V}_i$. We now set $C_u = A_{u,v_1} \cap ... \cap A_{u,v_n}$ and $D_u = B_{u,v_1} \cup ... \cup B_{u, v_n}$. We then observe that $C_u$ is an open neighborhood of $u$ in $X_{i+1}$, $D_u$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, and $\overline{C}_u \times \overline{D}_u \subseteq W$. Next we may find a finite collection $C_{u_1},...,C_{u_m}$ which covers $\overline{U}_i$. We may now set $E_{i+1} = C_{u_1} \cup ... \cup C_{u_m}$ and $F_{i+1} = D_{u_1} \cap ... \cap D_{u_m}$. Then $E_{i+1}$ is an open neighborhood of $\overline{U}_i$ in $X_{i+1}$, $F_{i+1}$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, $\overline{E}_{i+1}$ and $\overline{F}_{i+1}$ are both compact and $\overline{U}_i \times \overline{V}_i \subseteq \overline{E}_{i+1} \times \overline{F}_{i+1} \subseteq W_{i+1}$. Finally, since $X_i \longrightarrow X_{i+1}$ is a closed embedding we may let $U_{i+1}$ be the intersection of $E_{i+1}$ with some open subset of $X_{i+1}$ whose intersection with $X_i$ is $U_i$. Similarly we define $V_{i+1} \subseteq Y_{i+1}$. Then $U_{i+1} \cap X_i = U_i$ and $V_{i+1} \cap X_i = V_i$. Furthermore, $\overline{U}_{i+1}$ and $\overline{V}_{i+1}$ are compact and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W_{i+1}$, as desired.

To finish the proof we apply the procedure described repeatedly so that for every $j=i,i+1,...$ we have open subsets $U_j \subseteq X_j, V_j \subseteq Y_j$ such that $\overline{U}_j$ and $\overline{V}_j$ are compact and $\overline{U}_j \times \overline{V}_j \subseteq W_j$. Furthermore, by construction we have $U_{j'} \cap X_j = U_j$ and $V_{j'} \cap Y_j = V_j$ for every $j < j'$. Let $U = \cup_{j \geq i} U_j$ and $V = \cup_{j \geq i} V_j$. Then $U$ is an open subset of $colim_i X_i$, $V$ is an open subset of $colim_i Y_i$ and $(x,y) \in U \times V \subseteq g(W)$. This show that $g(W)$ is open, as desired.

  • $\begingroup$ Is there a reference for the case of CGWH-spaces? $\endgroup$
    – Tom
    Sep 7, 2015 at 12:33
  • $\begingroup$ See Peter May's book "Parametrized homotopy theory", Remark 1.6.4 (note that all limits and colimits mentioned there are taken in CGWH). $\endgroup$ Sep 7, 2015 at 13:31
  • $\begingroup$ Note also that even though the case discussed in Remark 1.6.4 is where the $Y_i$'s and $Z_i$'s are constant sequences, it actually implies the more general statement. $\endgroup$ Sep 7, 2015 at 13:41
  • $\begingroup$ It seems like the proof didn't use that the maps are closed embeddings but only that each $X_i\to X_{i+1}$ and $Y_i \to Y_{i+1}$ is an embedding, i.e. homeomorphism onto its (not necessarily closed) image, right? Also, the $Z_i$'s don't need to be locally compact but only Hausdorff and the $Y_i, Z_i$'s need a neighborhood basis of precompact opens, but don't need to be Hausdorff? Can a similiar statement be proven for filtered colimits? $\endgroup$ Dec 20, 2023 at 20:47

It is not an answer and I cannot let something wrong: the isomorphism is a general fact about locally finitely presentable categories. Let $\mathcal{K}$ be a locally presentable category. The pullback will preserve $\kappa$-filtered colimits if $\mathcal{K}$ is locally $\kappa$-presentable. But the category of $\Delta$-generated spaces is locally $\lambda$-presentable for $\lambda \geq 2^{\aleph_0}$. I had in mind that a pullback has $3$ elements which is of course not relevant here.

  • $\begingroup$ You need a locally finitely presentable category for this. An $\omega$-sequence is not $\kappa$-filtered for any uncountable regular cardinal $\kappa$. I know the category of $\Delta$-generated spaces is locally presentable, but I doubt it is locally finitely presentable. $\endgroup$
    – Zhen Lin
    Aug 28, 2015 at 15:02
  • $\begingroup$ You're right ! I had in mind the fact that a pullback contains 3 elements. The pullback preserves $\kappa$-filtered colimits with $\kappa$ a regular cardinal greater than $2^{\aleph_0}$. $\endgroup$ Aug 28, 2015 at 16:19
  • $\begingroup$ I'm not sure I believe that the category of $\Delta$-generated spaces is locally $\lambda$-presentable for $\lambda \ge 2^{\aleph_0}$. I might believe $\lambda > 2^{\aleph_0}$, but it's still something I haven't seen proved. $\endgroup$
    – Zhen Lin
    Aug 28, 2015 at 18:08
  • $\begingroup$ I am realizing by reading your answer that I don't know how to prove that this category is not locally finitely presentable. The only proof I know uses an axiomatization with formulae having $2^{\aleph_0}$ arguments. That does not prove that there does not exist any axiomatization of this category with formulae having a finite number of arguments. Of course, I doubt very much that this category is locally finitely presentable. $\endgroup$ Aug 28, 2015 at 19:05
  • 2
    $\begingroup$ I guess you are referring to the proof here? It doesn't really indicate what the axioms are supposed to be, but I gather that they are based on ultrafilter convergence in $\Delta^n$ or something like that. That is why I believe the category is locally $(2^{\aleph_0})^+$-presentable, but as I said, I haven't seen a complete proof. $\endgroup$
    – Zhen Lin
    Aug 28, 2015 at 19:36

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