I have two questions. For the first, we consider a projective morphism between two smooth, projective varieties over $k$, $f:X\rightarrow Y$. Let $\mathcal{L}$ be a line bundle and $f^*\mathcal{L}$ the pullback. Under what assumptions is $f*\mathcal{L}$ again a line bundle? (more generally what happens for a vector bundle $\mathcal{E}$ ?)
For the second, we consider a closed embedding of a smooth variety $X$ into a smooth and projective variety $Y$ , $j:X\rightarrow Y$ and $\mathcal{L}$ a line bundle on $X$. When is $j_*\mathcal{L}$ again a line bundle? (more generally what for a vector bundle $\mathcal{E}$ ?)
[Math] pullback and pushforward of line bundles
algebraic-geometry
Related Solutions
Pullback of invertible sheaves corresponds to pull-back of line bundles in the geometric sense.
In usual topology, a (complex) line bundle is a family of one-dimensional $\mathbb C$-vector spaces varying over a topological space $X$. More precisely, it is a topological space $V$, equipped with a continuous map $V \to X$, a zero section $X \to V$, a scalar multiplication action $\mathbb C \times V \to V$ compatible with the projections to $X$, an addition map $V \times_X V \to V$, again compatible with the projections to $X$, all satisfying some evident axioms, which more or less amount to requiring that these structure endow each fibre $V_x$ of $V$ over a point $x$ of $X$ with the structure of a one-dimensional $\mathbb C$-vector space, in such a way that $V$ is locally trivial.
Now if $\varphi: Y \to X$ is a continuous map, we may form the fibre product $\varphi^*V := V\times_X Y,$ and this is a line bundle over $Y$.
If you just think about how fibre product is defined, you see that the fibre of $\varphi^*V$ over a point $y \in Y$ is canonically identified with the fibre of $V$ over $\varphi(y)$. So if we think of $V$ as being the family $V_x$ of lines parameterized by the points $x \in X$, then $\varphi^*V$ is the family of lines $V_{\phi(y)}$ parameterized by $y \in Y$.
We may form the sheaf of sections $\mathcal L$ of $V$; this is a locally free sheaf of rank one over the sheaf of continuous $\mathbb C$-valued functions on $X$. The sheaf of sections of $\varphi^*V$ is then $\varphi^*\mathcal L$.
There is an exercise somewhere in Hartshorne which describes the analogue of the above theory of vector bundles in the algebro-geometric setting; everything goes over more-or-less the same way. I am now going to work in the algebro-geometric setting.
It is not so easy to relate the sections of $V$ to those of $\varphi^*V$; it depends very much on the space $Y$ and the morphism $\varphi$. For example, if $Y$ is just a point $x \in X$, then $\varphi^*V$ is just the line $V_x$, and its space of sections is one-dimensional (indepdently of what $V$ is). But it might be that $V$ has no sections that are non-zero at $x$.
More generally, a non-trivial bundle on $X$ might pull-back to something trivial on $Y$.
So you have to analyze this question on a case-by-case basis. Cohomology is one of the basic tools available here, but I am guessing that you're not at the point of using cohomology yet.
E.g. if $Y = \mathbb P^1$ and $\varphi: Y \to \mathbb P^2$ embeds $Y$ as a plane conic, then this image is a degree two curve, so $\mathcal O(1)$ pulls back to a degree two sheaf on $Y$, i.e. to $\mathcal O(2)$. Thus its space of global sections is three dimensional. So in this case $\varphi^*$ induces an isomorphism on spaces of global sections.
E.g. If $Y = \mathbb P^1$ but $\varphi: Y \to \mathbb P^2$ embeds it as a line, then $\varphi^*\mathcal O(1)$ is just $\mathcal O(1)$ on $\mathbb P^1$ (a line is a degree one curve), and so the map on sections is surjective, with a one-dimensional kernel (which is precisely the linear form cutting out the image of $\varphi$).
As Martin Brandenburg indicates in an answer, the sections $s_i$ give a projective embedding in the following concrete way: locally we may trivialize $V$, and so regard the $s_i$ as simply functions; we thus obtain a morphism $x \mapsto [s_0(x): \cdots : s_n(x)].$ Note that if we change trivialization, then the interpretation of the $s_i$ as functions changes by a nowhere zero function (the same function for all the $s_i$), which means that the point in $\mathbb P^n$ doesn't change. Thus the map is well-defined independent of the trivialization.
Finally, it might be helpful to know that in topology, the complex line bundles are classified (up to isomorphism) by homotopy classes of maps to $\mathbb C P^{\infty}$. Again, the map is given by pulling back $\mathcal O(1)$.
The situation in algebraic geometry is analogous to this, but more rigid: topological isomorphism is much less rigid than algebraic isomorphism (hence only the homotopy class of $\varphi$ matters), and all line bundles have lots of sections. In algebraic geometry, only sufficiently positive (i.e. sufficiently ample) bundles arise from maps to projective space.
In general, the answer is yes. Let's assume that $C$ is geometrically irreducible curve over any field $k$.
Note that we have a short exact sequence of sheaves
$$1\to \mathcal{O}_C^\times\to v_\ast \mathcal{O}_{C'}^\times\to (v_\ast\mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times\to 1$$
Taking the long exact sequence in cohomology gives
$$1\to k^\times\to k^\times\to ((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)(C)\to \mathrm{Pic}(C)\to \mathrm{Pic}(C')\to 1$$
where we note that $(v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times$ is finitely supported which is why its cohomology is zero. In particular, if $C$ has $n$ double points then it's not hard to see that
$$((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)(C)=\bigoplus_{x_i\text{ node}}((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)_{x_i}=(k^\times)^n$$
essentially because locally around a node $x_i$, with preimage points $p_1$ and $p_2$, we have that the curve has ring of functions $\{f(t)\in k[t]:f(p_1)=f(p_2)\}$ and locally around each of the two points $p_1$ and $p_2$ in $v^{-1}(x_i)$ we have that the functions just look $k[t]$ so the isomorphism $((v_\ast \mathcal{O}_{C'}^\times)/\mathcal{O}_C^\times)_{x_i}\to k^\times$ is something like $f\mapsto f(p_1)f(p_2)^{-1}$.
All in all we see that we have a short exact sequence
$$1\to (k^\times)^n\to\mathrm{Pic}(C)\to\mathrm{Pic}(C')\to 1$$
so that $\mathrm{Pic}(C)\to\mathrm{Pic}(C')$ is essentially never injective.
Best Answer
(Making my comment into an answer.)
The pullback of a vector bundle is always a vector bundle. The pushforward of a nontrivial vector bundle by a nontrivial embedding is never a vector bundle — it is trivial outside the image of the embedding.