Given an exact sequence of vector spaces $$\cdots\longrightarrow V_{i-1}\longrightarrow V_{i}\longrightarrow V_{i+1}\longrightarrow\cdots$$ I want to show that it is the same as having a collection of short ones such that
$$0\longrightarrow K_i \longrightarrow V_i \longrightarrow K_{i+1}\longrightarrow0$$
So to start I want to show exactness at an arbitrary $V_i$, so I space them suggestively:
$$\begin{array}{c} 0&\rightarrow &K_{i-1}&\rightarrow &V_{i-1}&\rightarrow &K_{i}&\rightarrow&0\\ &&&&0&\rightarrow&K_i&\rightarrow &V_i&\rightarrow&K_{i+1}&\rightarrow&0\\ &&&&&&&&0&\rightarrow&K_{i+1}&\rightarrow&V_{i+1}&\rightarrow&K_{i+2}&\rightarrow&0 \end{array}$$
I drop inclusions down among the corresponding $K_i$'s, and then compose until I get a function from $V_{i-1}$ to $V_i$ and one from $V_i$ to $V_{i+1}$. I check that the image of the first composite mess is the kernel of the second composite mess, which indeed it is.
Question: Am I done? Is showing exactness at one such $V_i$ enough? The question now prompts me to worry about the case were the orginal sequence isn't infinite in both directions...I'm not sure how that case is different?