$X\vee Y\cong (X\times\{y_0\})\cup (\{x_0\}\times Y)$

Solution 1:

Your map $f$ defined via $f_X$ and $f_Y$ is clearly continuous and induces a continuous bijection $\varphi : X \vee Y \to X\times\{y_0\}\cup \{x_0\}\times Y$.

Now let $U \subset X \vee Y$ be open. Then $q^{-1}(U)$ is open in $X \sqcup Y$ and can be written as $V_X \sqcup V_Y$ with open $V_X \subset X$ and open $V_Y \subset Y$. We have $$\varphi(U) = \phi(q(q^{-1}(U))) = f(q^{-1}(U)) = f(V_X \sqcup V_Y) = f_X(V_X) \cup f_Y(V_Y) \\= V_X\times\{y_0\}\cup \{x_0\}\times V_Y .$$ We shall show that this set is open in $ X\times\{y_0\}\cup \{x_0\}\times Y$. Let $* \in X \vee Y$ denote the common equivalence class of $x_0, y_0$.

Case 1 : $* \in U$. Then $x_0 \in V_X, y_0 \in V_Y$. The set $D = (V_X \times V_Y) \cap (X\times\{y_0\}\cup \{x_0\}\times Y)$ is open in $X\times\{y_0\}\cup \{x_0\}\times Y$. We have $$D = (V_X \times V_Y) \cap (X\times\{y_0\}) \cup (V_X \times V_Y) \cap (\{x_0\}\times Y) = V_X \times \{y_0\} \cup \{x_0\} \times V_Y .$$

Case 2: $* \notin U$. Then $x_0 \notin V_X, y_0 \notin V_Y$. The set $E = (V_X \times Y \cup X \times V_Y) \cap (X\times\{y_0\}\cup \{x_0\}\times Y)$ is open in $X\times\{y_0\}\cup \{x_0\}\times Y$. We have $$E = (V_X \times Y) \cap (X\times\{y_0\}) \cup (V_X \times Y) \cap (\{x_0\}\times Y) \\ \cup (X \times V_Y) \cap (X\times\{y_0\}) \cup (X \times V_Y) \cap ( \{x_0\}\times Y) \\= V_X \times \{y_0\} \cup \emptyset \cup \emptyset \cup \{x_0\} \times V_Y .$$

Remark:

The induced $\varphi$ is a bjection for any topology on $X \vee Y$. If $X \vee Y$ has the quotient topology with respect to $q$, then $\varphi$ is continuous. If we look closer to this part, we see that the proof does not require the continuity of $q$, but only the implication $q^{-1}(U)$ open $\Rightarrow U$ open.

Note that our above proof that $\varphi$ is an open map did not use that $X \vee Y$ has the quotient topology with respect to $q$, it works for any topology making $q$ continuous. But in fact the quotient topology is just the finest topology making $q$ continuous.