The following question has proved troublesome and prompted some deeper questions which I will elaborate on. Our definition of a left exact functor is one which takes exact sequences:

$$0 \rightarrow A \rightarrow B \rightarrow C$$

to an exact sequence.

Prove that a covariant functor $T: {_R}Mod → {_S}Mod$ is left exact if and only if exactness of

$$0 → A → B → C → 0~~~~~~~ (\star) $$

implies that

$$0 → TA → TB → TC ~~~~~~~(\dagger)$$

is exact.

$(\Leftarrow)$ (attempt)

Assume that the first sequence being exact implies that the second is, we now want to be able to replace $0$ in ($\dagger$) with $T(0)$ and replace the zero map with the image of the zero homomorphism under $T$, while retaining exactness of the sequence. This would be very easy if the image of $0$ under $T$ had to be $0$ (since then the map would also be zero) but I'm not sure if this is true. Either way we need the homomorphism to be the zero one since the second map in ($\dagger$) is injective.

I have seen similar results proved with initial and final objects, but we haven't actually covered them in the Category theory course I am taking. Also in other places where they are used we are normally dealing with isomorphisms of categories, not just functors.

($\Rightarrow$) This seemed like an easy direction but I have come across a possible complication. Assume that $T$ is left exact. Then the following sequence is exact:

$$T0 → TA → TB → TC $$

Let $f:A \rightarrow B$ be the map in ($\star$). We know that $f$ is injective and to have the result ($\dagger$) we need $T(f)$ to be injective and we are done.

I first did this as follows, if $f:A \rightarrow B$ is injective, then restricting the codomain we have an isomorphism $f:A \rightarrow f(A)$. Hence there exists $f^{-1} : f(A) \rightarrow A$ s.t. $ff^{-1} = 1_{f(A)}$ and $f^{-1}f = 1_{A}$. Since functors preserve identity maps we then have that $$T(f^{-1})T(f) = 1_{T(A)}$$

This appears to imply that $T(f)$ is injective, however we had to restrict $f$ to get this and so technically this is not the same function which normally isn't a problem but I feel like at a functorial level they might get mapped to completely different morphisms.

  • 2
    $\begingroup$ Proceed in several steps. First prove that, in either definition, $T 0 = 0$. Then show that $T$ preserves binary products and kernels (in both definitions). Finally, show that a functor that preserves $0$, binary products, and kernels is left exact (in both definitions). $\endgroup$ – Zhen Lin Apr 8 '14 at 11:11

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