# Doubt about the colimit of a sequence of topological spaces

Let $$X_1\stackrel{j_1^2}{\hookrightarrow }X_2 \stackrel{j_2^3}{\hookrightarrow}\dots$$ be a sequence of topological embeddings (i.e., $$j_n^{n+1}$$ is s sequence of injective continuous map such that $$j_n^{n+1}:X_n\hookrightarrow j_n^{n+1}(X_n)$$ is an homoemorphism). Its colimit is a topological space defined by: $$colim\, (X_n)=\bigsqcup_{n\in\mathbb N}X_n/\sim$$ where $$\sim$$ is the equivalence relation generated by $$x\sim j_n^{n+1}(x)$$ if $$x\in X_n$$.

I am trying to prove the following statement:

Let $$X_1\stackrel{j_1^2}{\hookrightarrow }X_2 \stackrel{j_2^3}{\hookrightarrow}\dots$$ be a sequence of topological embeddings such that $$X_n=X_{n_0}$$ for all $$n\geq n_0$$ and some $$n_0$$. Then $$colim\, (X_n)_n$$ is homeomorphic to $$X_{n_0}$$

It is easy to show that the map: $$\begin{array}{rclll} \phi:&colim\,(X_n)_n=\bigsqcup_{n=1}^{n_0}X_n/\sim&\longrightarrow X_{n_0}\\ &[(x,X_m)]&\longmapsto \phi\big([(x,X^m)]\big)=(j_{n}^{n+1}\circ \ldots \circ j_{n_0-1}^{n_0})(x) \end{array}$$ is continuous and bijective. How can I show that $$\phi$$ is also open?

Here's an extremely standard proof for dealing with colimits in general. It might seem rather long-winded, but it provides a lot of insight and is an extremely typical proof that you'll see repeated in many contexts.

The first thing to do is prove that this actually is a colimit according to the category theory definition, which I will provide.$$\DeclareMathOperator{colim}{colim}$$

Consider any sequence of topological spaces at all $$\{X_k\}_{k \in \mathbb{N}}$$, together with continuous $$\{j^k_m : X_k \to X_m\}_{k \leq m}$$, such that $$j^k_m \circ j^m_n = j^k_n$$ for all $$n \leq m \leq k$$ and such that $$j^k_k = 1_{X_k}$$ for all $$k$$.

Here, $$1_Y : Y \to Y$$ is the identity function.

We say that a colimit of $$X$$ (technically, a colimit of $$(X, j)$$) is a topological space $$A$$, together with continuous maps $$\{a_k : X_k \to A\}_{k \in \mathbb{N}}$$, such that

• For all $$m \leq k$$, we have $$a_k \circ j^k_m = a_m$$
• For all spaces $$Y$$ and all collections of continuous functions $$\{f_k : X_k \to Y\}_{k \in \mathbb{N}}$$, if for all $$m \leq k$$, $$f_k \circ j^k_m = f_m$$, then there exists a unique continuous function $$f : A \to Y$$ such that for all $$k$$, $$f \circ i_k = f_k$$.

From here, it's fairly straightforward to prove that

If $$(A, \{a_k : X_k \to A\}_{k \in \mathbb{N}})$$ and $$(B, \{b_k : X_k \to B\}_{k \in \mathbb{N}})$$ are both a colimit of $$X$$, then there exists a unique isomorphism (that is, a unique homeomorphism) $$g : A \to B$$ such that for all $$k$$, $$g \circ a_k = b_k$$.

To do this, come up with the unique continuous $$g : A \to B$$ such that for all $$k$$, $$g \circ a_k = b_k$$. Then, come up with continuous $$h : B \to A$$ such that for all $$k$$, $$h \circ b_k = a_k$$. Then, note that for all $$k$$, we have $$(g \circ h) \circ b_k = g \circ (h \circ b_k) = g \circ a_k = b_k = 1_B \circ b_k$$, where $$1_B$$ is the identity function. Conclude that by uniqueness, $$1_B = g \circ h$$. Do the same thing to conclude that $$1_A = h \circ g$$. Then $$g$$ and $$h$$ are inverse isomorphisms - that is, inverse homeomorphisms.

This is a special case of an extremely general theorem in category theory - the uniqueness of all kinds of colimits up to unique isomorphism, not just colimits indexed by $$\mathbb{N}$$. Note that almost no "topological" facts have been used in this proof at all. Indeed, the only topological facts which have been used are that composition of continuous functions is continuous, and that the identity function is continuous. So this same proof can apply in other situations - eg, consider a sequence of groups with group homomorphisms. This is the power of category theory.

Now, consider the case where the $$j^k_m$$ are inclusions. In particular, your example only starts with $$j^{k + 1}_k$$ for all $$k$$, but we can easily extend this to $$j^k_m$$ for all $$k \geq m$$ by composing inclusions (which is why there are two indices).

Now, you must prove that "the colimit" of $$X$$, as defined in your post, is actually "a colimit" of $$X$$, as defined above. The colimit consists not just of the space $$\colim(X) = \coprod\limits_{k \in \mathbb{N}} X_k / \sim$$, but also of the inclusion maps $$i_k : X_k \to \colim X$$ for all $$k$$. This is reasonably straightforward, and it's the only part of the proof where topology actually comes into play.

Now, consider the case where for all $$n \geq n_0$$, we have $$X_n = X_{n_0}$$ and $$j^n_{n_0} = 1_{X_{n_0}}$$. We can then prove that $$X_{n_0}$$ is a colimit of $$X$$ quite straightforwardly. We take $$\iota_k : X_k \to X_{n_0}$$ by $$\iota_k = j^{n_0}_k$$ if $$k \leq n_0$$, and $$1_{X_{n_0}}$$ if $$k \geq n_0$$. We then verify that this is indeed a colimit of $$X$$.

Finally, apply the uniqueness of colimits theorem to produce a homeomorphism $$\colim(X) \to X_{n_0}$$.

• Thanks for your answer! Everything is so much clearer to me now Commented Oct 15, 2021 at 8:20