Represent an isomorphism by $\leftrightarrow$.

HAVE exact sequence.

$$ 0 \rightarrow \mathbb Z \leftrightarrow \mathbb Z \rightarrow 0 $$


$$ \text{img} \left( 0 \rightarrow \mathbb Z \right) = 0 = \ker \left( \mathbb Z \leftrightarrow \mathbb Z \right)$$


$$ \text{img} \left( \mathbb Z \leftrightarrow \mathbb Z \right) = \mathbb Z = \ker \left( \mathbb Z \rightarrow 0\right) $$

NOW, make $\leftrightarrow$ the identity isomorphism. Rewrite the exact sequence as

$$ 0 \rightarrow \mathbb Z \rightarrow 0 $$


$$ \text{img} \left( 0 \rightarrow \mathbb Z \right) = \ker \left( \mathbb Z \rightarrow 0 \right) $$

but this would only hold if

$$ \mathbb Z = 0 $$

which is absurd.

The first sequence is intended to be translated from Hatcher's Algebraic Topology, page 114, corollary 2.14. However, this example of an isomorphism is ubiquitous when demonstrating exact sequences. I've seen it for months, but it hasn't been a problem until seeing Hatcher's proof. For clearer reference, his related example is his exact sequence

$$ 0 \rightarrow \tilde H_i \left( S^{n} \right) \leftrightarrow \tilde H_{i-1} \left( S^{n-1} \right) \rightarrow 0$$

I don't think the details of his proof are germane to my problem, but I referred it here in case I'm mistranslating what he wrote.

Clearly I'm doing something wrong, but I don't know how I'm tripping up, and I've wracked my brain over this for hours.

  • $\begingroup$ The rewritten exact sequence is not exact $\endgroup$
    – Rkb
    Commented Jun 9 at 8:32
  • $\begingroup$ @Rkb What distinguishes two isomorphic groups from being the same set/group/structure then? What form of equivalence can't be met for the sequence to fail to be exact? The identity map is an isomorphism, so what isomorphisms satisfy an exact sequence? Maybe there's a gap in my algebra I need plugged! $\endgroup$
    – Nate
    Commented Jun 9 at 8:33
  • 2
    $\begingroup$ You have presented a nice proof that you cannot collapse isomorphic subterms in exact sequences into one while preserving exactness :) If you let go of the erroneous belief that this is something you can do your problem vanishes. $\endgroup$ Commented Jun 9 at 13:00

1 Answer 1


The identity morphism also satisfies the exactness. Your first two observations are valid for all isomorphisms, in particular also the identity morphism. The error lies in translating the sequence $$0 \to \mathbb{Z} \overset{\mathrm{id}}{\to}\mathbb{Z} \to 0$$ to $$0 \to \mathbb{Z} \to 0$$ These sequences are just formally different in that, e.g., the second sequence only contains two morphisms whereas the first contains three morphisms.


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