Give the name $h$ to the mapping; i.e. $h(i,j)=(1,2)^i(1,2,3)^j$. The function h is __not__ a homomorphism. First h is a surjective map:

$h(1,0)=(1,2)$

$h(0,1)=(1,2,3)$

$h(2,0)=(1,2)^2=e\text{, the identity of }S_3$

$h(1,1)=(1,2)(1,2,3)=(2,3)$ (composition from right to left)

$h(0,2)=(1,2,3)^2=(1,3,2)$

$h(1,2)=(1,2)(1,2,3)^2=(1,2)(1,3,2)=(1,3)$

The above equations show that every element of $S_3$ is an image.

Fact: if h is a homomorphism from an abelian group G onto a group H, then H is abelian. You can try and prove this. So, if h were a homomorphism, then $S_3$ would be abelian, but it's not.

Alternatively, if h were a homomorphism,

$h(1,1)=h((1,0)+(0,1))=h(1,0)h(0,1)=(1,2)(1,2,3)$

$h(1,1)=h((0,1)+(1,0))=h(0,1)h(1,0)=(1,2,3)(1,2)$

But $(1,2)(1,2,3)\neq((1,2,3)(1,2)$