Lemma 102.70.1. There exists a system $G_1 \to G_2 \to G_3 \to \ldots $ of (abelian) topological groups such that $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological spaces is different from $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological groups.

## 102.70 Different colimit topologies

This example is [Example 1.2, page 553, TSH]. Let $G_ n = \mathbf{Q} \times \mathbf{R}^ n$, $n \geq 1$ seen as a topological group for addition endowed with the usual (Euclidean) topology. Consider the closed embeddings $G_ n \to G_{n + 1}$ mapping $(x_0, \ldots , x_ n)$ to $(x_0, \ldots , x_ n, 0)$. We claim that $G = \mathop{\mathrm{colim}}\nolimits G_ n$ endowed with the topology

is not a topological group.

To see this we consider the set

Using that $jx_0$ is never an integral multiple of $\pi /2$ as $\pi $ is not rational it is easy to show that $U \cap G_ n$ is open. Since $0 \in U$, if the topology above made $G$ into a topological group, then there would be an open neighbourhood $V \subset G$ of $0$ such that $V + V \subset U$. Then, for every $j \geq 0$ there would exist $\epsilon _ j > 0$ such that $(0, \ldots , 0, x_ j, 0, \ldots ) \in V$ for $|x_ j| < \epsilon _ j$. Since $V + V \subset U$ we would have

for $|x_0| < \epsilon _0$ and $|x_ j| < \epsilon _ j$. However, if we take $j$ large enough such that $j \epsilon _0 > \pi /2$, then we can choose $x_0 \in \mathbf{Q}$ such that $|\text{cos}(jx_0)|$ is smaller than $\epsilon _ j$, hence there exists an $x_ j$ with $|\text{cos}(jx_0)| < |x_ j| < \epsilon _ j$. This contradiction proves the claim.

**Proof.**
See discussion above.
$\square$

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