If $C$ is a chain complex of modules over a ring $R$ and $F: Mod_R \to Mod_S$ an additive exact functor then:

If $F$ is covariant then $H_n(FC)\cong FH_n(C)$.

Can anyone give my an idea or an idea to prove this statement? Thanks.

  • 3
    $\begingroup$ The exactness of the functor is key. Additive by itself won't have this property. You want to show kernels, images, and quotients are preserved. $\endgroup$
    – Zavosh
    Dec 8, 2014 at 2:24
  • $\begingroup$ But which is the sequence I need in order to compute the isomorphism? :/ @Prometheus $\endgroup$
    – Zill
    Dec 8, 2014 at 2:42
  • 2
    $\begingroup$ Use the definitions, the isomorphism comes out without any effort then. $\endgroup$ Dec 8, 2014 at 8:02

1 Answer 1


I just need this result and let me fill the proof as stated by @Zavosh in the comments: we'll prove image, kernel, and quotient is preserved under the exact functor.

Let's declare some notations, let $C=(C_n,\partial_n)_{n\in \mathbb N}$ be a complex, and $F$ be the exact covariant functor, with obtained complexes: $FC=(F(C_n),F(\partial_n))_{n\in \mathbb N}$ (This is a complex since $F$ is a functor and the composition of boundary maps factor through $F$.) Let $Z_n=\ker(\partial_n)\subset C_n$, and $B_n=Im(\partial_{n+1})\subset C_n$, and we want to show $$H_n(FC)\cong F(H_n(C))$$

We shall prove it in the following steps:

(i) Image is preserved by $F$

This only needs $F$ is a functor, since by a commutative diagram, we have $$Im(F(\partial_{n+1}))=F(\partial_{n+1})(F(C_{n+1})))=F(\partial_{n+1}(C_{n+1}))=F(Im(\partial_{n+1}))=F(B_n)\tag{1}$$

(ii) Kernel is preserved by $F$

Since the functor $F$ is exact, by applying $F$ to the short exact sequence:

$$0\to Z_{n}\xrightarrow{i_n} C_n\xrightarrow{\partial_n} B_{n-1}\to 0$$

we still get short exact sequence:

$$0\to F(Z_{n})\xrightarrow{F(i_n)} F(C_n)\xrightarrow{F(\partial_n)} F(B_{n-1})\to 0$$ The exactness tells you that

$\ker(F(\partial_n))\cong F(Z_n)=F(\ker(\partial_n)) \tag{2}$

(iii) Quotient is preserved by $F$

Applying $F$ to the Short exact sequence:

$$0\to B_{n}\xrightarrow{j_n} Z_n\xrightarrow{\pi_n} H_{n}(C)\to 0$$ where $j_n$ is the inclusion map, gives you $$0\to F(B_{n})\xrightarrow{F(j_n)} F(Z_n)\xrightarrow{F(\pi_n)} F(H_{n}(C))\to 0$$

The exactness tells you that $F(Z_n)/F(B_n)\cong F(H_n(C))\tag{3}$

(iv) Conclusion

Thus by steps above, we have $$H_n(FC)=\frac{\ker(F(\partial_n))}{Im(F(\partial_{n+1}))}\stackrel{by (1)(2)}{\cong}\frac{F(\ker(\partial_n))}{F(Im(\partial_{n+1}))}=\frac{F(Z_n)}{F(B_n)}\stackrel{by (3)}{\cong}F(H_n(C))$$

  • $\begingroup$ Sorry why the first step of the conclusion is correct by 1) and 2) ? I mean if we have for example quotients groups A/B and C/D with A iso to C and B iso to D then it is not true that A/B is iso to C/D (Just take Z/2Z and Z/3Z). Thank you $\endgroup$
    – Richard
    Jun 15, 2017 at 18:50
  • $\begingroup$ Moreover, to which commutative diagram are you referring to prove that the image is preserved? $\endgroup$
    – Richard
    Jun 15, 2017 at 18:52
  • $\begingroup$ Why $F(\partial_{n+1})(F(C_{n+1}))=F(\partial_{n+1}(C_{n+1}))$? $\endgroup$ May 24, 2018 at 20:42
  • $\begingroup$ Doesn't this use the fact that $F$ is exact and additive? To use the fact that $d_{n+1}:C_{n+1}\to B_n(C)$ is surjective, then so is $F(d_{n+1}):F(C_{n+1})\to F(B_n(C))$. And, therefore, $F(B_n(C))=Im F(d_{n+1})=B_n(F(C))$. $\endgroup$ May 24, 2018 at 21:03
  • 1
    $\begingroup$ @AndreGomes You can use that $0\to \operatorname{Ker}(d) \to C \to \operatorname{Im}(d) \to 0$ is a s.e.s.. So, applying $F$ gives another s.e.s. because $F$ is exact (which precisely says that $FC \to F(\operatorname{Im}(d))$. This doesn't require $F$ to be additive. $\endgroup$
    – MathsIsFun
    Oct 22, 2020 at 16:14

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