I agree, the answer is no in general. Let's take the usual bijection: rational maps from $X$ to projective space correspond to subspaces of the rational sections of line bundles on $X$.

Now you're looking for a way of embedding your variety $X$ into projective space so that when we represent the map $f$ in coordinates, $f(x) = (g_1(x) : \dots: g_n(x))$, for $g_i$ homogeneous of the same degree $d$. This means that the $g_i$ are sections of the special line bundle: $O(d)|_X$. So if there is a way of representing $f$ in this nice form, then the line bundle corresponding to $f$ must be of the special form $O(d)|_X$ for some embedding of $X$ into projective space.

Now from the Veronese embedding of $\mathbb{P}^n$ , finding such an $O(d)$ is the same as finding an $O(1)$. The punchline being that $f$ can be represented in this form if and only if $f$ comes from a line bundle $L$ where $L = O(1)|_X$ for some embedding of $X$ into projective space (i.e. $L$ is a very ample line bundle).

Now there are many examples of line bundles which are not very ample, and any map coming from those line bundles will not be able to be written in this form. For instance, if you take an elliptic curve, no matter how you embed it into projective space, a hyperplane will intersect it in more than one point. Thus, the line bundle coming from the divisor of a point on an elliptic curve gives a map that cannot be written in this form.

I agree, the answer is no in general. Let's take the usual bijection: rational maps from $X$ to projective space correspond to subspaces of the global sections of line bundles on $X$.

Now you're looking for a way of embedding your variety $X$ into projective space so that when we represent the map $f$ in coordinates, $f(x) = (g_1(x) : \dots: g_n(x))$, for $g_i$ homogeneous of the same degree $d$. This means that the $g_i$ are sections of the special line bundle: $O(d)|_X$. So if there is a way of representing $f$ in this nice form, then the line bundle corresponding to $f$ must be of the special form $O(d)|_X$ for some embedding of $X$ into projective space.

Now from the Veronese embedding of $\mathbb{P}^n$ , finding such an $O(d)$ is the same as finding an $O(1)$. The punchline being that $f$ can be represented in this form if and only if $f$ comes from a line bundle $L$ where $L = O(1)|_X$ for some embedding of $X$ into projective space (i.e. $L$ is a very ample line bundle).

Now there are many examples of line bundles which are not very ample, and any map coming from those line bundles will not be able to be written in this form. For instance, if you take an elliptic curve, no matter how you embed it into projective space, a hyperplane will intersect it in more than one point. Thus, the line bundle coming from the divisor of a point on an elliptic curve gives a map that cannot be written in this form.

I agree, the answer is no in general. Let's take the usual bijection: rational maps from $X$ to projective space correspond to subspaces of the global sections of line bundles on $X$.

Now you're looking for a way of embedding your variety $X$ into projective space so that when we represent the map $f$ in coordinates, $f(x) = (g_1(x) : \dots: g_n(x))$, for $g_i$ homogeneous of the same degree $d$. This means that the $g_i$ are sections of the special line bundle: $O(d)|_X$. So if there is a way of representing $f$ in this nice form, then the line bundle corresponding to $f$ must be of the special form $O(d)|_X$ for some embedding of $X$ into projective space.

Now from the Veronese embedding of $\mathbb{P}^n$ , finding such an $O(d)$ is the same as finding an $O(1)$. The punchline being that $f$ can be represented in this form if and only if $f$ comes from a line bundle $L$ where $L = O(1)|_X$ for some embedding of $X$ into projective space (i.e. $L$ is a very ample line bundle).

Now there are many examples of line bundles which are not very ample, and any map coming from those line bundles will not be able to be written in this form. For instance, if you take an elliptic curve, no matter how you embed it into projective space, a hyperplane will intersect it in more than one point. Thus, the line bundle coming from the divisor of a point on an elliptic curve gives a map that cannot be written in this form.

I agree, the answer is no in general. Let's take the usual bijection: rational maps from $X$ to projective space correspond to subspaces of the rational sections of line bundles on $X$.

Now you're looking for a way of embedding your variety $X$ into projective space so that when we represent the map $f$ in coordinates, $f(x) = (g_1(x) : \dots: g_n(x))$, for $g_i$ homogeneous of the same degree $d$. This means that the $g_i$ are sections of the special line bundle: $O(d)|_X$. So if there is a way of representing $f$ in this nice form, then the line bundle corresponding to $f$ must be of the special form $O(d)|_X$ for some embedding of $X$ into projective space.

Now from the Veronese embedding of $\mathbb{P}^n$ , finding such an $O(d)$ is the same as finding an $O(1)$. The punchline being that $f$ can be represented in this form if and only if $f$ comes from a line bundle $L$ where $L = O(1)|_X$ for some embedding of $X$ into projective space (i.e. $L$ is a very ample line bundle).

Now there are many examples of line bundles which are not very ample, and any map coming from those line bundles will not be able to be written in this form. For instance, if you take an elliptic curve, no matter how you embed it into projective space, a hyperplane will intersect it in more than one point. Thus, the line bundle coming from the divisor of a point on an elliptic curve gives a map that cannot be written in this form.