# $K$-homomorphisms for a field extension with a transcendental element

Suppose we have two extensions of $$K$$, namely $$L_1 = K(T)$$ ($$T$$ transcendental), and $$L_2$$ some other arbitrary extension. Then we look at the map $$\hom_K(L_1,L_2) \to L_2 : f \mapsto f(T)$$. First, I had to show that this is injective. I think the following calculation can show this. Take $$f,g \in \hom_K(L_1,L_2)$$ and suppose $$f(T) = g(T)$$. Let $$\sum_{i=0}^n a_iT^i \in L_1 = K[T]$$ arbitrary. Then $$f\left(\sum_{i=0}^n a_iT^i\right) = \sum_{i=0}^n a_if(T)^i = \sum_{i=0}^n a_ig(T)^i = g\left(\sum_{i=0}^n a_iT^i\right),$$ because $$a_i \in K$$ and $$f\mid_K = g\mid_K ={\rm id}$$. Hence $$f = g$$ on $$K[T]$$,and also on $$K(T)$$ since $$f(\frac{a}{b}) = f(a)/f(b)$$.

Now I am asked to describe the image of this map. Here I doubt. On the one hand, I guess it might be everything in $$L_2$$, since for a $$K$$-homomorphism we can send a transcendental element $$L \mapsto \alpha$$ to any element $$\alpha \in L_2$$, since $$T$$ does not have any algebraic relations to be satisfied. On the other hand, it also seems reasonable that its image is all the transcendental elements in $$L_2$$, since we must have $$T \mapsto S$$ a transcendental element in $$L_2$$? Anyone knows the image of this, and also how to prove it?

Let $$\phi : L_1 = K(T) \longrightarrow L_2$$ be a $$K$$-homomorphism of fields. If $$\phi(T) = x \in L_2$$, then Im$$(\phi) = K(x)$$. Since a non-trivial field homomorphism is an isomorphism onto the image, we have that $$x$$ is transcendental over $$K$$.
Let $$\sigma : \text{hom}_K (L_1, L_2) \longrightarrow L_2$$. Then $$\sigma(\phi) = x$$. Thus Im $$(\sigma)$$ is the set of all transcendental elements of $$L \setminus K$$.