Suppose $f: M \to N$ and the map $g: N \to M$, where $M$ and $N$ are $R$-mdoules. We wish to find the limit over the diagram $$M \rightleftarrows N$$ in the two cases: (1) Where $fg = 1_N$ and $gf = 1_M$; (2) When $fg$ and $gf$ are not necessarily the identity on their respective modules.
Assuming that $f \circ g = 1_N$ and $g \circ f = 1_M$, then I've tried a few candidates for the limit object, call it $L$. The problem is that I can never seem to get the limit cone to commute.
If $L = M/\operatorname{im} g \oplus N/\operatorname{im} f$, and the legs of the cone are the canonical projection maps $\pi_1, \pi_2$, then for this to actually commute we must have $$f \circ \pi_1 (m + \operatorname{im} g \oplus n + \operatorname{im} f) = f(m) = \pi_2 (n)$$ where the right-hand side is only in the image of $f$ if in fact $n$ is zero and hence $m$ is in the kernel of $f$. This gives me the feeling that the limit is in fact the zero-module, the terminal object. But this does not seem to satisfy the commutativity condition for the legs of the cones.
I also tried $M/\ker f \oplus N/ \ker g$ and even $\operatorname{ker} f/ \operatorname{im} g \oplus \ker g/ \operatorname{im} f$.
I am not having any luck, and have spent many hours thinking about this problem and I suspect the answer is somewhat trivial.
The limit is the subset of pairs $(m,n)$ of $M\times N$ satisfying $f(m)=n$ and $g(n)=m$.
When $fg=1_N$, one of the conditions becomes redundant, and we just require $g(n)=m$ for all pairs $(m,n)$, so the limit is $\{(g(n),n)|n\in N\}\cong N$.
Likewise if $gf=1_M$, the limit is just $M$.