# Topology of wedge products

I have a question about the quotient topology induced on the wedge sum $S^{\,2} \vee S^1$, (where $S^n$ denotes the unit sphere in $\mathbb{R}^n$). In this topological space, the subsets $S^1$ and $S^2$ are not open, is that correct ?

Because, if we denote by $p$ the quotient map that sends the disjoint union $S^1 \amalg S^2$ to $S^{\,2} \vee S^1$, then for example the inverse image of $S^1$ under this map is $S^1 \cup \{x_0\}$, where $\{x_0 \} \subset S^2$ denotes the point in $S^2$ that is indentified to the intersection point by $p$.

But this set $S^1 \cup \{x_0\}$ is not open in $S^1 \amalg S^2$, is that right ?

The reason I am asking is because I don't quite understand an argument by Hatcher in his book on algebraic topology, on page 47, where he uses van Kampen's theorem to compute the fundamental group $\pi_1(S^{\,2} \vee S^1)$. The part that I am struggeling with is that van Kampen's theorem (as given on page 43 of the same book) refers to a space $X$ that is the union of open sets.

Now if I set $X = S^{\,2} \vee S^1$ then (if I am right above) I am struggeling to understand how I can can apply van Kampen's theorem, if I am right and $S^2$ and $S^1$ are not open sets.

What am I doing wrong?

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$S^1$ is not open in $S^1 \vee S^2$, but $S^1$ is a strong deformation retract of an open subspace of $S^1 \vee S^2$ ... now apply Seifert van Kampen. – martini Apr 15 '12 at 13:49
@martini: Ah ok, I just found an explanation by Hatcher himself, also referring to what you recommend. Thanks for your comment ! – harlekin Apr 15 '12 at 14:12