Formalizing the Wedge Sum as "glueing" spaces together So, in John Lee's Introduction to Topological Manifolds, he states that $\mathbb{R} \vee \mathbb{R} \cong $ the union of the $x$ and $y$ axes in $\mathbb{R}^2$. I am trying to formalize this.
Suppose we let $X_1 = (\mathbb{R}, 0)$ (basepoint $0$) and let $X_2 = (\mathbb{R}, 0)$. So, I'm trying to show $\mathbb{R} \vee \mathbb{R} \cong (\mathbb{R} \times \{0\}) \cup (\{0\} \times \mathbb{R}) =: D$
Having chosen the basepoints as I have, $\mathbb{R} \vee \mathbb{R} = X_1 \sqcup X_2 / \{(0,1), (0,2)\}$ where $(0,1)$ and $(0,2)$ are just the images of $0$ under the canonical injections $j_i : X_i \to X_1 \sqcup X_2$. 
It seems to me then that the right homeomorphism candidate is $f: D \to \mathbb{R} \vee \mathbb{R}$ sending $(0,0)$ to $\{(0,1), (0,2)\}$, $(x,0)$ to $\{(x,1)\}$ for $x \neq 0$ and $(0,x)$ to $\{(x,2)\}$ for $x \neq 0$. I think it's pretty clear that $f$ will be a bijection. However, the problem then arises in showing the continuity of $f$. In particular, we need to know what topology is on $\mathbb{R} \vee \mathbb{R}$. My first thought was to simply endow $\mathbb{R} \vee \mathbb{R}$ with the quotient topology induced by $f$. Then, showing $f$ is a homeomorphism would be quite easy. However, we may also see $\mathbb{R} \vee \mathbb{R}$ as the quotient space of $X_1 \sqcup X_2$ obtained by collapsing $\{(0,1), (0,2)\}$ to a point. In this light, $\mathbb{R} \vee \mathbb{R}$ is a partition of $X_1 \sqcup X_2$ , so there is some equivalence relation $\sim$ on $X_1 \sqcup X_2$ such that $\mathbb{R} \vee \mathbb{R} = X_1 \sqcup X_2 / \sim$. Then, the topology on $\mathbb{R} \vee \mathbb{R}$ is actually the quotient topology induced by the map $q: X_1 \sqcup X_2 \to X_1 \sqcup X_2 / \sim$ sending $x$ in the disjoint union to its equivalence class with respect to $\sim$. If this is the topology we are using, I imagine showing $f$ is a homeomorphism would probably be much more difficult.
Which topology do we choose in such a situation?
 A: The topology of the wedge sum is by definition the quotient topology.
The theory can be simplified a little bit: Let $(X,x_0)$ and $(Y,y_0)$ be pointed topological spaces, $X\sqcup Y$ the disjoint union, and $X\lor Y$ the wedge sum (with basepoints $x_0$ and $y_0$). Let $\pi\colon X\sqcup Y\to X\lor Y$ be the quotient map. Then a subset $A\subseteq X\lor Y$ is open if and only if $\pi^{-1}(A)$ is open. This is equivalent to saying that $\left\{x\in X:\pi(x)\in A\right\}$ ad $\left\{y\in Y:\pi(y)\in A\right\}$ are open in $X$ and $Y$, respectively. (I'm identifying elements of $X$ and $Y$ with their images in $X\sqcup Y$).
Since we are dealing with a wedge sum of $\mathbb{R}$ with itself, we can simplify the notation a bit: Given $x\in\mathbb{R}$, let $x_1$ be the class of $(x,0)$ in $\mathbb{R}\lor\mathbb{R}$ and $x_2$ be the class of $(0,x)$ in $\mathbb{R}\lor\mathbb{R}$. In this manner, $x_i$ is just the representative of $x$ in the $i$-th copy of $\mathbb{R}$. In particular, we have $0_1=0_2$.
Let $\pi\colon\mathbb{R}\sqcup\mathbb{R}\to\mathbb{R}\lor\mathbb{R}$ be the quotient map. Now let $A\subseteq\mathbb{R}\lor\mathbb{R}$ be open. This means that $\pi^{-1}(A)$ is open, so both sets $A_i=\left\{x\in\mathbb{R}:x_i\in A\right\}$ ($i=1,2$) are open in $\mathbb{R}$. The definition of $f$ yields $f^{-1}(A)=A_1\times\left\{0\right\}\cup\left\{0\right\}\times A_2$, which you can show to be open in $D$.

For example, if we have $x\in A_1\setminus\left\{0\right\}$, then $A_1\setminus\left\{0\right\}$ is open in $\mathbb{R}$, so $(A_1\setminus\left\{0\right\})\times\mathbb{R}$ is open in $\mathbb{R}^2$ and thus
  $$((A_1\setminus\left\{0\right\})\times\mathbb{R})\cap D=(A_1\setminus\left\{0\right\})\times\left\{0\right\}$$
  is open in $D$, contains $(x,0)$ and is contained in $f^{-1}(A)$. This proves that every nonzero point of $A_1\times\left\{0\right\}$ is interior to $f^{-1}(A)$. Similarly, every nonzero point of $\left\{0\right\}\times A_2$ is interior to $f^{-1}(A)$.
The only problem is to prove that $(0,0)$, if it belongs to $f^{-1}(A)$, is interior to $f^{-1}(A)$. But notice that $f(0,0)=0_1=0_2$, so $(0,0)\in f^{-1}(A)$ if and only if $0\in A_1$ and $0\in A_2$. In this case, $(A_1\times A_2)\cap D$ is an open neighbourhood of $(0,0)$ in $D$, which is therefore interior.

This shows that $f$ is continuous. To prove that $f$ is a homeomorphism, we can find its inverse: Let $g\colon\mathbb{R}\sqcup\mathbb{R}\to D$ be given by $g(x,1)=(x,0)$ and $g(x,2)=(0,x)$. Then $g(0,1)=(0,0)=g(0,2)$, so $g$ factors uniquely through a map $G\colon\mathbb{R}\lor\mathbb{R}\to D$, that is, to a map $G$ such that $g=G\circ\pi$. Since $g$ is continuous, then $G$ is also continuous (this is the universal property of quotients: If we have a quotient map $q\colon X\to X/\sim$, then a function $G\colon X/\sim\to Y$ is continuous if and only if $G\circ q$ is continuous).
It is easy to see that $G=f^{-1}$, so $f^{-1}$ is continuous.
