Is the bijection shown by commutativity? "Let $G_1. G_2$ be groups. Prove that the map
$$ \varphi: G_1 \times G_2 \rightarrow G_2 \times G_1$$
$$ \varphi: (g_1, g_2) \mapsto (g_2, g_1)$$
defines an isomorphism between $G_1 \times G_2$ and $G_2 \times G_1$.
I can prove the homomorphism bit but in the answers, it just says "Obviously $\varphi$ is bijective". Why is this obvious? Is it because it is commutative?
 A: What is the kernel of $\phi$? Indeed, if $(x,y)\in\ker(\phi)$ where $x\in G_1,y\in G_2$ then $$\phi(x,y)=(y,x)=(e_{G_2},e_{G_1})\Longrightarrow y=e_{G_2},x=e_{G_1}$$ and so the map is a monomorphism (because you showed it is a homomorphism befor). Moreover by taking $(y,x)\in G_2\times G_1$ you have $(x,y)$ in $G_1\times G_2$ and so the map is surjective as well.
A: If $\varphi(a,b)=(0,0)$ then $a=0$ and $b=0$ so the kernel is trivial and thereby $\varphi$ injective. Now for surjectivity for any $(a,b) \in G_2 \times G_1$ we have that $\varphi(b,a)=(a,b)$. 
A: For injectivity, observe that if $\varphi(g_1,g_2)=\varphi(g_1',g_2')$, then by the defintion of $\varphi$, we get $(g_2,g_1)=(g_2',g_1')$. Hence $g_1=g_1'$ and $g_2=g_2'$ and $\varphi$ is injective.
For surjectively, we ask if for any $(g_2,g_1)\in G_2\times G_1$ we could find some $(a,b)\in G_1\times G_2$ such that $\varphi(a,b)=(g_2,g_1)$. But the answer is clearly affirmative because $\varphi(g_1,g_2)=(g_2,g_1)$.
A: The application $\psi$ is defined by:
$$\psi: G_2 \times G_1 \rightarrow G_1 \times G_2$$
$$\psi: (g_1,g_2) \mapsto (g_2,g_1)$$
We have:
$$\varphi \circ \psi = id_{G_2 \times G_1}$$
$$\psi \circ \varphi = id_{G_1 \times G_2}$$
Therefore $\varphi$ and $\psi$ are bijective and reciprocate.
