Let $X$ be the line with two origins, the result of identifying two lines except their origins. Let $X_n$ be the result of identifying two lines except their intervals $(-\frac{1}{n},\frac{1}{n})$. $X_n$ is a Hausdorff space that exists in $\mathbb{R}^2$ and is homotopic to a circle, and $X_{n+1}$ is naturally a quotient space of $X_n$. I like to imagine that we have two zipper sliders gradually approaching each other without touching, which results in $X$. Is this intuition correct, i.e., is $X$ is the direct limit of $X_1\rightarrow X_2\rightarrow\cdots$, where the maps are quotient maps? I think I was able to show that if we have a directed diagram of quotient maps, the natural map from each space $X_i$ to the direct limit $X$ is also quotient map, and in our case this should imply the direct limit is same as line with two origins.

If so, can we use this to get the homology and fundemental groups of $X$? I know that homology commutes with direct limits for nice spaces, such as increasing union of CW complexes, but I saw an example that this is not true in general; also we are having quotient maps but not inclusion here. I am aware of this question; just curious if the groups can be obtained by some abstract nonsense.


1 Answer 1


Yes, $X$ is the direct limit of this sequence. This is essentially immediate from the universal property of quotient spaces: a map out of the direct limit is just a map out of $\mathbb{R}\sqcup \mathbb{R}$ which coincides on the two copies outside of $(-1/n,1/n)$ for all $n$, and that just means it coincides everywhere except $0$.

You can deduce the homology and homotopy groups of $X$ from the following pair of facts, which immediately imply that $H_*$ and $\pi_*$ preserve the direct limit:

Let $K$ be a compact Hausdorff space. Then:

  1. Any continuous map $f:K\to X$ lifts to a continuous map $K\to X_n$ for some $n$.
  2. If $g,h:K\to X_m$ are continuous maps whose compositions with $X_m\to X$ are equal, then there is some $n\geq m$ such that their compositions with $X_m\to X_n$ are equal.

To prove (1), let $0$ and $0'$ be the two origins in $X$ and let $A=f^{-1}(\{0\})$ and $A'=f^{-1}(\{0'\})$. Then $A$ and $A'$ are disjoint closed subsets of $K$, so they have disjoint open neighborhoods $U$ and $U'$. Note that then $\partial U$ is closed and disjoint from $A$ and $A'$, and so $f(\partial U)$ is a compact subset of $X\setminus\{0,0'\}$. This means there is some $n$ such that $f(\partial U)\cap [-1/n,1/n]=\emptyset$. Now lift $f$ to $X_n$ by mapping it to the first copy of $\mathbb{R}$ on $U$ and the second copy of $\mathbb{R}$ on $K\setminus U$. This is obviously continuous on $U$ and on the interior of $K\setminus U$. On $\partial U$, it is continuous since the two copies of $\mathbb{R}$ are identified in $X_n$ on a neighborhood of the image of $\partial U$.

To prove (2), let $f:K\to X$ be the common composition of $g$ and $h$ with $X_n\to X$ and let $A=f^{-1}(\{0,0'\})$. Note that then $g$ and $h$ must coincide in a neighborhood $U$ of $A$: if $g(x)=h(x)=0$, then there is a neighborhood of $x$ on which both $g$ and $h$ map into the first copy of $\mathbb{R}$, and so must coincide, and similarly if $g(x)=h(x)=0'$. By compactness of $K\setminus U$, $U$ must contain $f^{-1}([-1/n,1/n])$ for some $n\geq m$. Since $g$ and $h$ coincide on $U$, this means that their compositions with $X_m\to X_n$ are equal.


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