I'm reading Advanced Modern Algebra: Part 1 by Rotman and I'm having trouble verifying the functor in the following statement is left exact:
kernels of R-maps are pullbacks. Thus, kernels are inverse limits. Therefore, if an additive contravariant functor $F : RMod → SMod$ preserves inverse limits, it preserves kernels in particular, and so it is left exact.
I understand, from other answers on math stack exchange, the case for covariant functors but am struggling to see the exact calculation required to prove it directly in the contravariant case.
If we have (where $p =0$-map) $$\require{AMScd} \begin{CD} A@>f>>B@>g>>C@>p>>0\\ \end{CD}$$ then if this is exact we have $ker(p)\cong im(g)\cong C$ by surjectivity of g and $ker(g)\cong im(f)$ and applying the contravariant functor gets: $$\require{AMScd} \begin{CD} 0@>F(p)>>F(C)@>F(g)>>F(B)@>F(f)>>F(A)\\ \end{CD}$$ so for this to be exact we need $ker(F(g))\cong im(F(p))$ and $ker(F(f))\cong im(F(g))$.
What I understand preserve kernels to mean is that $F(ker(p))\cong ker(F(p))$ , $F(ker(g))\cong ker(F(g))$, $F(ker(f))\cong ker(F(f))$. So then $F(ker(p))\cong F(C)\cong ker(F(p))\cong 0$.
However already this doesn't quite look right to me and, if it is correct, I'm not sure how to proceed in proving injectivity of F(f). Any help clearing this up for me would be appreciated.