general topology - 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?