A particular exact sequence of vector spaces using full rank linear maps

let $V_a,V_b,V_c$ and $V_d$ be vector spaces where $$dim(V_a) = n-1$$ $$dim(V_b)=n$$ $$dim(V_c)=n$$ $$dim(V_d)=n+1$$

Let the following all be full rank linear maps:

$$gamma_{ab}: V_a rightarrow V_b$$
$$gamma_{ac}: V_a rightarrow V_c$$
$$gamma_{bd}: V_b rightarrow V_d$$
$$gamma_{cd}: V_c rightarrow V_d$$

Furthermore, say that the following sequence is exact: (sorry I don’t know how to make the mappings ABOVE the arrows…)

$$V_a overset{(gamma_{ab},gamma_{ac})}{longrightarrow} V_b oplus V_c overset{gamma_{ba}-gamma_{ca}}{longrightarrow} V_d$$

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I would like to learn more about this situation in any way possible. Here are some things that I believe can be said (please correct me if I’m wrong):

The dimensions of $V_b$ and $V_c$ are both one higher than the dimension of $V_a$; thus, since $gamma_{ab}$ and $gamma_{ac}$ are full rank maps, they inject $V_a$ into $V_b$ and $V_c$ respectively; thus $im(gamma_{ab}) cong V_a$ and $im(gamma_{ac}) cong V_a$.

By the isomorphism theorem, we have $frac{V_b oplus V_c}{ker(gamma_{ca}-gamma_{ba})} cong im(gamma_{ba}-gamma_{ca}) < V_d$

Since this sequence is exact, we have that $ker(gamma_{ca}-gamma_{ba}) cong im(gamma_{ab},gamma_{ac})$. Although in general when invoking the isomorphism theorem with respect to some algebraic structure, $A cong B$ and $C cong D$ does not imply that $frac{A}{C} cong frac{B}{D}$, in our situation, since isomorphism classes of vector spaces are completely determined by dimension, we can conclude that:

$$frac{V_b oplus V_c}{V_a oplus V_a} cong frac{V_b oplus V_c}{im(gamma_{ab},gamma_{ac}) } cong frac{V_b oplus V_c}{ker(gamma_{ca}-gamma_{ba})} cong im(gamma_{ba}-gamma_{ca}) < V_d$$

Now,
$$dimleft(frac{V_b oplus V_c}{V_a oplus V_a}right)=dim(V_b)+dim(V_c)-2dim(V_a)=n+1+n+1-2n=2$$ and thus $dim(im(gamma_{ba}-gamma_{ca}))=2$.

Is all of this true? What else that can (or can’t) be said?

I know this is a pretty open ended questions, but it seems my open ended questions on this site get some really insightful answers… Sometimes you don’t know what you are looking for until you find it!

Thanks