2 Clarified statement; added 21 characters in body
1. I don't know offhand, would have to look up. The standard reference is Arbarello, Cornalba, Griffiths, Harris, Geometry of Algebraic Curves, Springer. Great book.

2. Yes.

3. For $d=1$ (and $g>0$), the map is an isomorphism to its image. In general, a fiber of $\alpha_d$ is a projective space consisting of effective divisors in the same linear equivalence class.

Hyperellipticity plays no role when $d=1$. For $d=2$, if the curve is not hyperelliptic, the map is an isomorphism (again to its image if $g > 1$). If the curve is hyperelliptic, the map is injective outside the set of divisors of the form $(x,y)+(x,-y)$ and crushes this set to one point.

Edit: Clarified point raised in comments. Isomorphism to its image.

1
1. I don't know offhand, would have to look up. The standard reference is Arbarello, Cornalba, Griffiths, Harris, Geometry of Algebraic Curves, Springer. Great book.

2. Yes.

3. For $d=1$ (and $g>0$), the map is an isomorphism. In general, a fiber of $\alpha_d$ is a projective space consisting of effective divisors in the same linear equivalence class.

Hyperellipticity plays no role when $d=1$. For $d=2$, if the curve is not hyperelliptic, the map is an isomorphism. If the curve is hyperelliptic, the map is injective outside the set of divisors of the form $(x,y)+(x,-y)$ and crushes this set to one point.