Identify a pullback line bundle on $\mathbb{P}^1$ Consider a degree $d$ map $f:\mathbb{P}^1 \to \mathbb{P}^m$ for $d \geq 1$ and $m \geq 2$, together with a line bundle $\mathcal{O}_{\mathbb{P}^m}(l)$ over $\mathbb{P}^m$ for $l \geq 1$. Then we have a pullback line bundle $f^*\mathcal{O}_{\mathbb{P}^m}(l)$ over $\mathbb{P}^1$.
Is $f^*\mathcal{O}_{\mathbb{P}^m}(l) \cong \mathcal{O}_{\mathbb{P}^1}(dl)$ always true?
Let $[x_0,x_1]$ and $[y_0, \dots, y_m]$ be the homogeneous coordinates of $\mathbb{P}^1$ and $\mathbb{P}^m$, respectively. Then one can argue that a degree $l$ homogeneous polynomial in $y_0, \dots, y_m$ pullbacks to a degree $dl$ homogeneous polynomial in $x_0$ and $x_1$. But these may not span the global sections $H^0 (\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(dl))$. For example, consider $m = 2$ and $l=3$, and a degree $2$ map $f:\mathbb{P}^1 \to \mathbb{P}^2$ defined by
$$[x_0,x_1] \mapsto [x_0^2,x_1^2,0].$$
Then the global section $x_0^5x_1$ in $H^0 (\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(6))$ is missing from the pullbacks of degree $3$ polynomials in $y_0,y_1,y_2$.
What's wrong with the above argument?
 A: As was pointed out by Sasha, the pull back map need not be an isomorphism on global sections.
If you take $E$ a vector bundle on a scheme $X$ and a map $f: Y\to X$ you get a map $f^*: H^0(X, E)\to H^0(Y, f^*E)$ which is not an isomorphism in general (for instance take $X=\mathbb{P}^n$ and $Y=\mathbb{A}^n$ an affine subspace, or $Y=x$ a point).
To prove the mentionned claim:
$$f^*\mathcal O(\ell)\simeq \mathcal{O}(d\ell)$$
you may notice that it suffices to prove it for $\ell=1$ as $f^*(E\otimes F)\simeq f^*E\otimes f^*F$.
Then you have the following proposition (which is in Hartshorne for instance):
Assume that $X$ is an $R$-scheme, giving a morphism $f: X\to \mathbb{P}_R^n$ is the same as giving a line bundle $\mathcal{L}$ over $X$ and $(n+1)$ sections $s_0,...,s_n$ of $\mathcal{L}$ generating the fiber at each point via the correspondance $\mathcal{L} \leftrightarrow f^*\mathcal{O}(1)$.
The proof is simply to notice that $X$ is covered by open subsets $U_i=\{x\in X|s_i(x)\neq 0\}$, which may not be affine, but for any affine $V$ in $U_i$ you get a morphism $V\to V_i\simeq \mathbb{A}^n$ (viewed as the complement of the hyperplane $x_i=0$ in $\mathbb{P}^n)$ given by $(s_0/s_i,...,s_n/s_i)$, these morphisms all agree and thus you get a well defined morphism $U_i\to V_i$, which again agree on overlaps and you get a well defined morphism $X\to \mathbb{P}^n$.
You also get a morphism $\mathcal{L}\to f^*(\mathcal{O}(1))$ sending, over $U_i$ the section $s_i$ to $x_i$, these maps by definition glue together to give you an isomorphism of line bundles.
For more details you can look up in Hartshorne.
Now if you take a morphism $f: \mathbb{P}^1\to \mathbb{P}^m$ given by $(m+1)$ homogenous polynomial of degree $d$ thus by $m+1$ sections of $\mathcal{O}(d)$ you get by the mentionned claim applying the proposition.
A small note: here I assumed that you took $f$ directly given by homogenous polynomials of degree $d$, depending on your definition of degree you may use a slightly different approach (like the projection formula if your definition comes from intersection theory).
In any case in view of the aforementionned proposition and the fact a line bundle over $\mathbb{P}^1$ if always of the form $\mathcal{O}(d)$ for some $d$, a morphism of degree $d$ from $\mathbb{P}^1$ to $\mathbb{P}^m$ will always be given by sections of $\mathcal{O}(d)$ i.e by $m+1$ homogenous polynomials of degree $d$.
