Fix brackets

Author: GeoffreySangston

spaces/S000107/properties/P000229.md

@@ -9,15 +9,15 @@ refs:
   name: Answer to "(Certain) colimit and product in category of topological spaces"
 ---
 
-Since [S107|P86], it is enough to show that the path-component of $0$
-is [P229]. By {{mathse:5012784}}, this component equals 
+Since {S107|P86}, it is enough to show that the path-component of $0$
+is {P229}. By {{mathse:5012784}}, this component equals 
 $\mathbb{R}^\infty = \{y : y^n = 0\text{ for all but finitely many }n\}$. We argue that $\mathbb{R}^\infty$ is
-contractible, since [P199|P229].
+contractible, since {P199|P229}.
 
 The claim follows once we argue that $F : \mathbb{R}^\infty \times [0, 1] \to \mathbb{R}^\infty$, $(x, t) \mapsto tx$, is continuous.
 By {{mathse:3961052}}, the subspace topology on $\mathbb{R}^\infty$ coincides with the weak topology,
 where a set $U \subset \mathbb{R}^\infty$ is open if and only if $U \cap \mathbb{R}^n$ is open for
 each $\mathbb{R}^n := \{x \in \mathbb{R}^\infty : x^m = 0\text{ if } m > n\}$. By {{mathse:833227}},
 the product of $\mathbb{R}^\infty \times [0, 1]$ also has the weak topology, where again a set $U \subset \mathbb{R}^\infty \times [0, 1]$ is open if and only if $U \cap (\mathbb{R}^n \times [0, 1])$ is open for each $n$.
-Then because the restrictions of $F$ to each $\mathbb{R}^n \times [0, 1]$ are continuous, it follows
+Then because the restrictions of $F$ to each $\mathbb{R}^n \times [0, 1]$ are continuous and valued in $\mathbb{R}^n$, it follows
 that $F$ is continuous.