Suppose I have a (group) homomorphism $f: SU(2) \to SO(3)$.
In my book (Symmetry and Quantum Mechanics by Corry) I have completed the following two exercises:
Exercise 2.36. Show that the set of 3 × 3 orthogonal matrices that may be joined to the identity matrix by a continuous path in O(3) is the subgroup of rotations SO(3). (Hint: the determinant is a polynomial in the entries of a matrix, hence continuous.)
Exercise 2.37. Prove that the kernel of f : SU(2) → SO(3) is the set of matrices in SU(2) that commute with all of the Pauli matrices $σ_1,σ_2,σ_3$. Then prove that the only such matrices in $GL(2, \mathbb{C})$ are scalar matrices (i.e., multiples of the identity). Conclude that the kernel of f is {$\pm I$}.
Now the next line in my book says that
The previous two exercises demonstrate that $f$ induces an isomorphism between SU(2)/{±I} and a subgroup of SO(3), namely the image of $f$.
I do not follow this argument. My understanding is as follows. We are defining some new function $g$ which is the restriction of $f$ to $SU(2)/\{±I\}$ and stating that $g$ is an isomorphism between $SU(2)/\{±I\}$ and $\textrm{Im}(g)$ (incidentally, I have diverged here from Corry; I think his saying "namely the image of $f$" is an error and it should be "namely the image of $f$ less the identity in $SO(3)$ since we have excluded $\ker{f}$ from our domain"). That $g$ is surjective is immediate from the definition of the codomain. That $g$ is injective is not obvious to me (maybe I can work hard to prove it, but my question is how does this follow from the aforementioned two exercises in particular?). I know that injectivity is equivalent to the kernel of the map containing only $I$ in the domain, but we don't have this for $f$ so this route doesn't work. It seems to me like we're somehow saying that we removed the only case of noninjectivity by removing $\pm I$ from the domain, but this strikes me as strange.
Finally, it seems to me that by isomorphism here we must mean an isomorphism between sets (since the domain of $g$ is not a group), right (this is my other question)?
Thank you for any help you can give me!
$G=SU(2)/\{\pm 1\}$ is not a subset of $SU(2),$ so there is no such thing as a restriction of a map on $SU(2)$ to this group. Further, $G$ is very much a group. The image of a homomorphism is the quotient of the domain by the kernel (this is one of (the first?) isomorphism theorems of group theory). So, I would suggest going back to group theory before quantum mechanics.