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Let $(X,x_0),(Y,y_0),(Z,z_0)$ be based spaces. Define the wedge sum $X\vee Y\vee Z$ to be $X\sqcup Y\sqcup Z$ modulo $x_0\sim y_0\sim z_0$. How does this wedge sum relates to the subspace $$S=\{(x,y,z)\in X\times Y\times Z \;|\; \text{only one of the entries is not the base point} \}$$

I Know they are identified but i'm not sure how to prove it. Let $$f:X\sqcup Y\sqcup Z\to S;\; f(x)=(x,y_0,z_0), f(y)=(x_0,y,z_0), f(z)=(x_0,y_0,z)$$ Then obviously $f$ is surjective and such that $f(x_0)=f(y_0)=f(z_0)$ hence $f$ facotors through $X\vee Y\vee Z$ to give a homeomorphism. Is this argument correct and especially is $f$ continuous?

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It is enough to consider just two spaces $X,Y$. Then it follows from formal nonsense (check the details!) that we have a continuous bijection $X \vee Y \to S := \{(x,y) \in X \times Y : x = x_0 \vee y = y_0\}$ mapping $x \mapsto (x,y_0)$ and $y \mapsto (x_0,y)$. It seems to be not so trivial that this is a homeomorphism. One has to show that this bijection maps open sets to open sets. If $U \subseteq X \vee Y$ is open, distinguish two cases: (a) $U$ contains the base point. (b) $U$ does not contain the base point.

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