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Suppose now that the objects in question are abelian topological groups $G$ so that morphisms are continuous group homomorphisms. Given an exact sequence of abelian topological groups $0 \to G''\to G \to G''$ is the functor $\textrm{Hom}(H,-)$ a left-exact functor? $H$ is an abelian topological group.

We know that if we only care about group homomorphisms then $\textrm{Hom}(H,-)$ is left-exact, but will the condition of continuity now change anything? It seems there is a problem because to prove exactness at $\textrm{Hom}(H,G)$ requires defining a map from $H \to G''$ which may not be continuous.

Edit: PinkElephants has shown us that this functor may fail to be exact. However if we consider a specific case which is the exact sequence

$$0 \to \Bbb{Z} \to \Bbb{R} \stackrel{e^{2\pi i x}}{\longrightarrow } S^1 \to 0$$

then is the functor $\textrm{Hom}(S^1, -)$ left-exact? What about the contravariant functor $\textrm{Hom}(-,S^1)$, is it still right-exact?

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If $\textrm{Hom}(G, -)$ fails to be left exact then you have defined exact sequence incorrectly. –  Zhen Lin Dec 10 '12 at 23:35
    
Continuity, in any case, cannot solve anything, for you always put the discrete topology on groups! –  Mariano Suárez-Alvarez Dec 10 '12 at 23:36
    
@MarianoSuárez-Alvarez What if I consider $S^1$ as a topological group the topology on here is the Euclidean topology? –  user38268 Dec 10 '12 at 23:38
    
@ZhenLin I made an edit above. –  user38268 Dec 10 '12 at 23:43
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Let me just mention the amazing fact that if you consider only compact abelian groups, then this category actually turns out to be abelian. In particular, Hom(S^1, -) is left-exact as long as you restrict to compact groups. –  Piotr Pstrągowski Dec 11 '12 at 2:23
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2 Answers 2

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$\DeclareMathOperator{Hom}{Hom}$Given a short sequence $$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0,\tag{$\ast$}$$ in an additive category, the sequence of abelian groups $$ 0 \to \Hom(G,A) \xrightarrow{f_\ast} \Hom(G,B) \xrightarrow{g_\ast} \Hom(G,C) $$ is exact for all $G$ if and only if $f$ is a kernel of $g$ and, dually, $$ 0 \to \Hom(C,G) \xrightarrow{g^\ast} \Hom(B,G) \xrightarrow{f^\ast} \Hom(A,G) $$ is exact for all $G$ if and only if $g$ is a cokernel of $f$.

This can be interpreted as follows: Since we want $\Hom$ to be left exact, the minimal requirement that a short sequence $(\ast)$ in an additive category should satisfy in order to deserve to be called exact is that $f$ be a kernel of $g$ and that $g$ be a cokernel of $f$.

From this point of view, the example in Pink Elephant's answer is not a good one: the identity $G \to H$ is a monomorphism and an epimorphism, but it is not an isomorphism and a fortiori neither a kernel nor a cokernel.

The trouble of calling all kernel-cokernel sequences $(\ast)$ exact is that they lack the closure conditions you need in order to prove the diagram lemmas necessary for homological algebra. A particularly convenient (hence convincing) approach to exact sequences (albeit not intrinsic as in abelian categories) is usually attributed to Quillen, see exact categories. In a nutshell what this means is that you have to think about what sequences you want to call exact.

Outside of the additive setting exactness can be dealt with in various ways but it gets even more complicated. See e.g. Borceux-Bourn's book.

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Let $H$ be an abelian topological group whose topology is not discrete, and let $G$ have the same underlying group as $H$ but with the discrete topology. Consider the short exact sequence $0\to G\to H\to 0\to 0$ where $G\to G$ is the identity on sets.

Now the sequence $0\to \mathrm{Hom}(H,G)\to \mathrm{Hom}(H,H)\to \mathrm{Hom}(H,0)$ is not exact at $\mathrm{Hom}(H,H)$, as $\mathrm{Id}_H$ is in the kernel but not the image.

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Thanks for your answer. However if the topologies are all inhereted from the euclidean topology (e.g. $G''= \Bbb{Z}$, $G = \Bbb{R}$ and $G'= S^1$) then will the functor be left-exact? –  user38268 Dec 11 '12 at 0:31
    
You may see my edit in the question above for further queries. –  user38268 Dec 11 '12 at 0:34
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