I have to work with the definition (for attaching an n-cell to a topological space X) to be the push-out of the following diagram : 
Now, The question :
Prove that $S^2 × S^2$ is obtained by attaching a 4-cell to $S^2∨S^2$
My attempt:
So I have the following diagram :
My aim : Consider $X$ to the push-out of the diagram and then show that it's homeomorphic to $S^2 \times S^2$ .
From $S^2∨S^2$ to $S^2 \times S^2$ we take the inclusion.
And consider, $g : D^4 \to S^2 \times S^2$ by, $g(x_1,x_2,x_3,x_4)=((\frac{x_1}{\sqrt{{x_1}^2+{x_2}^2}},\frac{x_2}{\sqrt{{x_1}^2+{x_2}^2}}),(\frac{x_3}{\sqrt{{x_3}^2+{x_4}^2} },\frac{x_4}{\sqrt{{x_3}^2+{x_4}^2}} ))$
Then, by definition of Push-out we have that $\exists$ continuous map $h : X \to S^2 \times S^2$. Commuting diagram gives that $h$ must be surjective (since, $g$ is clearly surjective).
Now, if I can show that $h$ is injective, then we are done , since $X$ is compact and $S^2 \times S^2$ is Hausdorff.
So how to prove that $h$ is injective?
Thanks in advance for help.
Here's an outline:
Let $p: D^2 \to S^2$ be the obvious projection collapsing the boundary $D^2$ to a marked point point $*$. Then write $D^4 = D^2 \times D^2$. Embed $S^2 \vee S^2 \subset S^2 \times S^2$ as $(S^2 \times \{*\}) \cup (\{*\} \times S^2)$. The map you want to use is $g(x,y) \to (p(x),p(y))$ for $(x,y) \in D^2 \times D^2$. It is easy to show this is injective on the interior of $D^4$. For example, if $x \not = x'$ are two distinct points in the interior of $D^2$, then $p(x) \not = p(x')$ so $g(x,y) \not = g(x',y')$. You can also show that the images of $D^4$ and $S^2 \vee S^2$ only intersect in the images of $S^3$, so these are identified by the pushout, and injectivity of $h$ will follow.