# Wedge Sum Embedding with Inclusions

Let $X$ and $Y$ be two disjoint topological spaces, $x_0\in X$, $y_0\in Y$ and we consider the Wedge Sum (the quotient of the union by the relation $x_0\sim y_0$).

I want to proof that $\pi \circ i_X$ and $\pi \circ i_Y$ are embedding, with $\pi:(X+Y) \rightarrow (X\vee Y)$ the canonical projection.

I have already proved that they are injective and continuous. The problem is to prove that it is closed (or open or $(\pi \circ i_X)^{-1}:(\pi \circ i_X)(X) \rightarrow X$ continuous).

-Closed: I take $F$ closed in $X$.

• If $x_0 \not\in F$, no problem, $(\pi \circ i_X)(F)=F"$
• If $x_0 \in F$, I get $(\pi\circ i_X)(F)=\{[x] \mid x \in F, x\neq x_0\}\cup\{[x_0]\}=A$. That subset is closed if and only if $\pi^{-1}(A)$ is closed in $X+Y$. But $\pi^{-1}(A)=\pi^{-1}( \{[x] \mid x \in F, x\neq x_0\}\cup\{[x_0]\})$, that is, $\pi^{-1}( \{[x] \mid x \in F, x\neq x_0\})\cup \pi^{-1}(\{[x_0]\})=F\cup\{y_0\}$.

$F\cup\{y_0\}$ is closed if and only if $(F\cup\{y_0\})\cap X$ and $(F\cup\{y_0\})\cap Y$ are closed. $(F\cup\{y_0\})\cap X$ is closed by hypothesis, but $(F\cup\{y_0\})\cap Y=\{y_0\}$ because $X\cap Y=\emptyset$.

But $\{y_0\}$ does not have to be closed in Y.

Where is the problem?

• an embedding isn't closed in general, so there's no problem – Exodd Dec 14 '14 at 20:25

Let $X$, $Y$ be pointed spaces, then $X\vee Y = (X+Y)/x_0\sim y_0$ is a pointed space with basepoint $*=[x_0]=[y_0]$. The inclusion $X\hookrightarrow X+Y$ induces an injective continuous map $i\colon X\to X\vee Y$. Now the continuous map \begin{align*} X + Y &\longrightarrow X, \\ x &\longmapsto x \,\,\text{ for $x\in X$},\\ y &\longmapsto x_0 \text{ for $y\in Y$} \end{align*} induces a map $q\colon X\vee Y\to X$. Since $q\circ i = \operatorname{id}_X$ we can conclude that $i$ is a homeomorphism onto its image.
(In you terminology $(\pi\circ i_X)^{-1}=q\big|_{\operatorname{im}(i)}$, so we have a continuous inverse.)
an embedding is an homeomorphism with the image, but in general isn't closed. Given $F\subseteq X$, you have to prove that $\pi\circ i_X(F)$ is closed in the induced topology on $\pi\circ i_X(X)$. That means you have to look for a closed subset $C\subseteq X\vee Y$ such that $$C\cap \pi\circ i_X(X)=\pi\circ i_X(F)$$
And it is easy to prove that $$C=\pi\circ i_Y(Y)\cup \pi\circ i_X(F)$$ Is the desired closed set
• If I am not mistaken you are saying that an if F is closed that $(\pi\circ i_X)(F)$ must be closed in $(\pi\circ i_X)(X)$ but not closed in $X\vee Y$ – user201185 Dec 14 '14 at 21:05