I'm having a bit of difficulty with exercise 5.16 from Rotman's An Introduction to Homological Algebra (second edition).
The exercise (at least the relevant part) reads
Prove that every left exact covariant functor $T:{}_R\mathbf{Mod} \to \mathbf{Ab}$ preserves pullbacks.
Now, I went back and checked the definition, and a covariant functor is defined to be left exact if
$$ 0\to A\to B\to C$$ being exact implies
$$ 0\to TA \to TB\to TC$$
is exact, with no mention of additivity.
Now, I'm aware of this question, but the comment on the question doesn't apply given the definition in Rotman, and the accepted answer to the question doesn't get me any further than the work I've already done. Thus this new question.
My work, and a precise statement of exactly where I'm running into trouble without additivity
It's not hard to show that left exact functors preserve direct sums,
and it's also not hard to check that
$$\require{AMScd}\begin{CD}
D @>>> C
\\
@VVV @VVgV
\\
B @>>f> A
\end{CD}$$
is a pullback square if and only if
$$0 \to D \to B\oplus C \newcommand\toby\xrightarrow \toby{f\pi_B-g\pi_C} A $$
is exact.
Then if we apply $T$ to this exact sequence, we get the exact sequence
$$ 0 \to TD \to TB \oplus TC \toby{T(f\pi_B-g\pi_C)} TA. $$
Now when $T$ is additive, we have
$$T(f\pi_B -g\pi_C) = (Tf)\pi_{TB}-(Tg)\pi_{TC},$$
so this exact sequence is the correct exact sequence to show that
$$\begin{CD}
TD @>>> TC
\\
@VVV @VVTgV
\\
TB @>>Tf> TA
\end{CD}
$$
is a pullback square.
However if $T$ is not additive, then it is not clear that the exact sequence we get from applying $T$ to the exact sequence from the pullback square in ${}_R\mathbf{Mod}$ is the correct exact sequence to prove that $TD$ is the pullback in $\mathbf{Ab}$. This is where I am stuck.
The question
Hence my questions:
- Is there a way to finish the proof without using additivity?
- Is there something special about these particular categories?
- Is there perhaps a counterexample, showing that we need to assume additivity? (I'm not aware of any left exact but not additive functors (nor is Google, based on a quick search), so I don't even know where to begin searching for a counterexample)
Best Answer
Roland left an excellent comment reminding me that the addition in an Abelian category (well technically category with finite biproducts) can be defined purely in terms of the direct sum, diagonal and codiagonal maps.
If $f,g : A\to B$ are morphisms, we define $f+g$ to be the composite $$ A\newcommand\toby\xrightarrow\toby{\Delta_A} A \oplus A\toby{f\oplus g}B\oplus B \toby{\nabla_B} B,$$ where $\Delta_A : A\to A\oplus A$ is the diagonal map and $\nabla_B : B\oplus B\to B$ is the codiagonal.
Then clearly any functor which satisfies $T(f\oplus g)=Tf\oplus Tg$ and $T\Delta_A = \Delta_{TA}$ and $T\nabla_B = \nabla_{TB}$ is additive.
Then as long as $T$ preserves the biproduct (meaning $T(A\oplus B)=TA\oplus TB$ and $T\iota_X=\iota_{TX}$ and $T\pi_X = \pi_{TX}$ for the structure morphisms $\iota_A,\iota_B,\pi_A,$ and $\pi_B$), and $T$ preserves $0$, meaning that $T0 = 0$, $T$ has these properties. (Equality here means up to natural isomorphism.)
This is because $\Delta_A$ is the map such that $\pi_{A,i}\Delta_A=\newcommand\id{\operatorname{id}}\id_A$ for $i=1,2$, and then taking $T$ of this, we have that $T\Delta_A$ satisfies $\pi_{TA,i}T(\Delta_A)=\id_{TA}$ for $i=1,2$, thus $T\Delta_A=\Delta_{TA}$. A dual argument shows that $T\nabla_B=\nabla_{TB}$, and finally $f\oplus g$ is defined to be the map such that $$\pi_{B,1}(f\oplus g)\iota_{A,1} = f,$$ $$\pi_{B,2}(f\oplus g)\iota_{A,2} = g,$$ $$\pi_{B,1}(f\oplus g)\iota_{A,2} = 0,$$ and $$\pi_{B,2}(f\oplus g)\iota_{A,1} = 0.$$ Again, it's not hard to see that the preservation of 0 and the biproduct ensures that $T(f\oplus g)$ satisfies the requirement to be $Tf\oplus Tg$.
Finally, if $T$ is left exact, then $T$ preserves $0$ and the biproduct, and thus is additive
(As indicated in my question, I've already proved this, but I figured I'd add it for any future readers)
$T$ preserves $0$
Depending on how you interpret the definition, this is either definitional, (since you can read the definition as implying that you take $T$ of the entire left exact sequence to get the resulting exact sequence) or must be proved, if you take it to mean only that extending the sequence $TA\to TB \to TC$ by zero on the left results in an exact sequence, but it turns out that there's no difference.
Consider $$ 0 \to 0 \toby{\id_0} 0 \toby{\id_0} 0.$$ This is clearly exact, so $$ 0 \to T0 \toby{\id_{T0}} T0 \toby{\id_{T0}} T0$$ is exact, which implies that $$0=\ker \id_{T0} = T0=\operatorname{im}_{\id_{T0}},$$ as desired.
$T$ preserves the biproduct
Proof:
We know $\pi_A\iota_A =\id_A$, so $(T\pi_A)(T\iota_A)=\id_{TA}$. This also applies to $\pi_B\iota_B$, so $T\pi_B$ must be an epimorphism. Hence, exactness of $$0\to A \toby{\iota_A} A\oplus B \toby{\pi_B} B \to 0$$ not only implies exactness of $$0\to TA \toby{T\iota_A} T(A\oplus B) \toby{T\pi_B} TB, $$ but actually the exactness of $$0\to TA \toby{T\iota_A} T(A\oplus B) \toby{T\pi_B} TB \to 0,$$ and the fact that $(T\pi_A)(T\iota_A)=\id_{TA}$ implies that this exact sequence is split, giving $T(A\oplus B) \simeq TA\oplus TB$.