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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?

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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 $L \setminus K$.

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