Do injectivity and surjectivity imply short exact sequence

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In Vakil's The Rising Sea (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf), (2.1.1.1) shows that because $0\longrightarrow m_p\longrightarrow O_p\longrightarrow \mathbb{R}\longrightarrow 0$ is a short exact sequence, $O_p/m_p\simeq \mathbb{R}$.

What I am questioning is how it was known that the sequence was an exact sequence. We knew beforehand that $0\longrightarrow m_p\longrightarrow O_p$ and $O_p\longrightarrow \mathbb{R}\longrightarrow 0$ are exact (injectivity and surjectivity). Does this automatically imply that the above "short sequence" is in fact a short exact sequence? If so, how? Or is other information required to prove a sequence is a short exact sequence?

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Pavel's comment basically answers this, but the thing to keep in mind in general is that for any surjective map $f:A \to B$ in an abelian category (e.g. vector spaces over $\mathbb R$), there is always a short exact sequence $$ 0 \to \ker f \to A \to B \to 0, $$

and for $f:A \to B$ injective a short exact sequence $$ 0 \to A \to B \to \operatorname{coker} f\to 0. $$

And to simplify the story, both of these sequences are special cases of what we get for $f$ arbitrary, $$ 0 \to \ker f \to A \to B \to \operatorname{coker} f\to 0. $$