Proving that isomorphism of field extension is an equivalence relation I'm reading Stewart's Galois Theory Third Edition. In Chapter 4 he gives the definition of the isomorphism of field extension as follows: 

An isomorphism between two field extension $l : K \rightarrow \hat{K}$, $j: L \rightarrow \hat{L}$ is a pair $(\lambda, \mu)$ of field isomorphisms $\lambda: K \rightarrow L, \mu : \hat{K} \rightarrow \hat{L}$ such that $j(\lambda(k)) = \mu(l(k))$ for all $k \in K$.

Then in the exercise he asks to prove that this isomorphism of field extensions is an equivalence relation. The reflectivity and transitivity are easy to see for me but I'm not sure if I'm dealing with symmetry correctly. 
I say (keep the notations as in the definition): 

If $\lambda \sim \mu$ then we have $j(\lambda(K)) = \mu(l(K))$, then there exists $j^{-1}$, $l^{-1}$ such that $j^{-1}(\mu(\hat{K})) = \lambda(l^{-1}(\hat{K}))$ which gives $\mu \sim \lambda$. 

However, the inverse of a field extension hasn't defined yet so I'm not sure if this is legit. Or maybe I'm just going the wrong track? I appreciate any help! 
 A: You want to invert $\lambda$ and $\mu$, not the extensions (these generally aren't invertible...). In more detail:
You want to show that $l \sim j$ implies $j \sim l$, that is, you wand to find a pair that gives you an isomorphism between the extensions $j$ and $l$ given that you already have such a pair $(\lambda,\mu)$ between $l$ and $j$. I claim that $(\lambda^{-1},\mu^{-1})$ is a pair that yields an isomorphism between $j$ and $l$. We have to show that $$l(\lambda^{-1}(l)) = \mu^{-1}(j(l)) \, \text{ for all } l \in L.$$ This is equivalent to $$\mu(l(\lambda^{-1}(l)) = j(l),$$
and since $(\lambda,\mu)$ is an isomorphism of field extensions between $l$ and $j$ (i.e., we are assuming from that start that $l \sim j$), this is equivalent to $$ j(\lambda(\lambda^{-1}(l)) = j(l),$$ which is obviously true for all $l \in L$.
A: A field extension is a triple $(K,l,K_1)$. We say that this is isomorphic to $(L,j,L_1)$ if we have an isomorphism $(\lambda,\mu)$ which is composed by two maps since we need the map between the base fields, and the map between the extended fields, and we want that the diagram commutes. We want to show that the field extensions are divided into equivalent classes (not the isomorphisms). Of course a field extension is isomorphic to itself since you can chose $(\lambda,\mu) = (Id_K,Id_{K_1})$ to get riflessivity. If you have $(K,l,K_1) \simeq (L,j,L_1)$ with isomorphism $(\lambda_1,\mu_1)$ and $(L,j,L_1) \simeq (H,t,H_1)$ with isomorphism $(\lambda_2,\mu_2)$ you can take $(\lambda,\mu) = (\lambda_2\lambda_1,\mu_2\mu_1)$ to get transitivity. If you want simmetry just take $(\lambda^{-1},\mu^{-1})$.
