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Define the wedge sum of two pointed spaces $X$ and $Y$ as $$ X\vee Y := X\sqcup Y / \sim $$ where $\sim$ is the equivalence relation which identifies the base points of $X$ and $Y$. I have read in some lecture notes that the canonical map $$X\vee Y \to X\times Y $$ is a homoemorphism onto its image $X\times \{y_0\} \cup \{x_0\} \times Y$. However, I do not see why the inverse map $$ X\times \{y_0\} \cup \{x_0\} \times Y \to X\vee Y $$ is continuous. It is given by $$ (x,y_0) \mapsto [x], \quad (x_0,y) \mapsto [y], $$ so it restricts to continuous maps on $X\times \{y_0\}$ and $\{x_0\} \times Y$. If $X$ and $Y$ are $T_1$-spaces, then these subspaces would be closed and the map would be continuous on all of $X\times \{y_0\} \cup \{x_0\} \times Y$. But how do I proceed if $X$ or $Y$ is not a $T_1$-space?

Many thanks in advance!

Simba
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1 Answers1

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You don't. For example, if $X$ and $Y$ are discrete, the closed subspace $X \subset X \vee Y$ is mapped to $X \times \{y_0\},$ which is not closed, because the product topology on $X \times Y$ is also discrete. Thus we'd better work in a category of, say, weakly Hausdorff spaces, or require in the definition of a pointed space $X$ that $* \to X$ is a closed map. I don't see any mention of that in these notes, or in the previous lecture, which makes me think that it's a mistake.

Арсений Кряжев
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