Refering to Serge Lang's "Algebra" pp. 133-134 on Abelian Categories, the following is unclear to me. Let $Q$ be an additive category and $F\stackrel{f}\rightarrow E$ a morphism. Let $A \stackrel{\alpha}\rightarrow F$ and $B \stackrel{\beta}\rightarrow F$ be two category-theoretic kernels of $f$ (as defined in the text). It is claimed that $\alpha$ and $\beta$ are isomorphic in a suitable category. What is this category? It is natural to consider the category of morphisms of $Q$ of the form $X \rightarrow F$, where $X$ is some object of $Q$. But i still can't see why $A$ and $B$ are isomorphic.


Lemma: If $\alpha:A\rightarrow F$ is a kernel of $f:F\rightarrow E$, then $\alpha \circ f = 0$.

Proof: By the definition of the kernel, we have exactness of the sequence:

$$0 \rightarrow \operatorname{Hom}(A,A)\rightarrow \operatorname{Hom}(A,F) \rightarrow \operatorname{Hom}(A,E)$$

But $\alpha$ is the image of ${id}_A$ above, so alpha must go to zero in $\operatorname{Hom}(A,E)$.

Now, given two kernels, $\alpha$ and $\beta$ we have the exact sequence:

$$0 \rightarrow \operatorname{Hom}(B,A)\rightarrow \operatorname{Hom}(B,F) \rightarrow \operatorname{Hom}(B,E)$$

But $\beta\in\operatorname{Hom}(B,F)$, is a kernel, so we know that the image of $\beta$ in $\operatorname{Hom}(B,E)$ is zero, so $\beta = \alpha \circ i$ for some $i$ in $\operatorname{Hom}(B,A)$.

Similarly, $\alpha = \beta \circ j$ for some $j\in \operatorname{Hom}(A,B)$.

Now let $k = i \circ j$. Then $k \in \operatorname{Hom}(A,A)$. But since $\alpha \circ k = \alpha \circ i \circ j = \alpha$, and $\operatorname{Hom}(A,A)\rightarrow \operatorname{Hom}(A,F)$ is 1-1, this means that $k = {id}_A$.

Similarly, $j \circ i = {id}_B$.

So $\alpha$ and $\beta$ are isomorphic in the category of maps in $Q$ which terminate in $F$.

  • $\begingroup$ Just curious: Lang does define kernels this way, not via the unique factorization property? Moreover, it may be worth mentioning that a kernel is necessarily a monomorphism (you of course proved this implicitly). $\endgroup$ – t.b. Jun 9 '11 at 1:14
  • $\begingroup$ Yes, I looked it up in Google Books. books.google.com/… $\endgroup$ – Thomas Andrews Jun 9 '11 at 1:17
  • $\begingroup$ Thanks! Google doesn't like me recently and never lets me look at the page I want to look at... $\endgroup$ – t.b. Jun 9 '11 at 1:18
  • $\begingroup$ Thank you for your nice answer. Two induced questions please: 1) Why do you compose the morphisms from left to right? 2) I still can't see why a kernel is a monomorphism, not even what that means in an abstract abelian category. Why is the map from Hom(A,A) to Hom(A,E) injective? $\endgroup$ – Manos Jun 10 '11 at 3:23
  • $\begingroup$ One answer per comment: The map from Hom(A,A) to Hom(A,E) is injective because the sequence $0\rightarrow Hom(A,A)\rightarrow Hom(A,E)$ is exact, by the definition of kernel where $X=A.$ $\endgroup$ – Thomas Andrews Jun 10 '11 at 3:49

Thomas's proof can be shortened by using properties of a universal object (here a terminal object), as Lang suggests. Consider a category $K$ (from “kernel”) where objects are morphisms $\gamma$ of $Q$ such that $f\circ\gamma=0$ and where morphisms are commuting triangles as in slice categories.

Take an arbitrary kernel $\alpha$. By Thomas's lemma, $\alpha$ is an object of $K$. For every object $\beta$ of $K$, let $B$ be its domain, $\beta\in\operatorname{Hom}(B,F)$, $f\circ\beta=0$, then $\beta$ maps to $0$ in the sequence

$$\operatorname{Hom}(B,A) \to\operatorname{Hom}(B,F) \to \operatorname{Hom}(B,E)$$

then, by exactness, $\beta$ has a preimage in this sequence, i.e. $\beta$ factorizes into $\alpha$. $\operatorname{Hom}(B,A)\to\operatorname{Hom}(B,F)$ in this sequence is injective, then $\beta$ factorizes uniquely. Then $\beta$ factorizes uniquely into $\alpha$. Then $\alpha$ is terminal in $K$. Terminal objects are isomorphic by the unique isomorphism.


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