Things I was surprised to learn are false

OK, so this one is not very interesting, but I could have sworn it was true and believing it unquestioningly led me to scratch my head for about an hour one day when I used it in a proof of something else. The correct statement is that a graph has no odd cycles if and only if it has 0 or 1 vertices or is bipartite¹.

Specifically, I thought the following was true:

If \(F\) is a contravariant functor from the (weak) homotopy category of topological spaces \(\text{Ho}(\text{Top}_*)\) to the category \(\text{Set}_∗\) of pointed sets which sends coproducts (i.e. wedges) to products and weak/homotopy pushouts (or CW triads) to weak pullbacks, then \(F\) is representable.

The theorem is true for the homotopy category of connected pointed spaces, but connectedness is necessary. I learned about this from Karol Szumilo’s answer to a MathOverflow question by Michael Shulman. Note that Brown representability also fails for unpointed spaces.

It is true that \(\lim^1\) vanishes for an inverse system of Abelian groups satisfying the Mittag-Leffler condition, but I thought that this was true not only for Abelian groups but in any Abelian category satisfying Grothendieck’s AB4 (i.e., cocomplete and where the coproduct of monomorphisms is a monomorphism).

This time the misunderstanding is not totally my fault, the generalization to AB4 categories was claimed with an erroneous proof in a paper by Jan-Erik Roos from 1961². In 2002, Neeman and Deligne constructed a counterexample and in 2006 Roos published a paper proving that \(\lim^1\) does vanish for an inverse system satisfying the Mittag-Leffler condition in any Abelian category that satisfies AB3 (i.e., is cocomplete), AB4* (i.e., is complete and the product of epimorphisms is an epimorphism) and has a set of generators.

There are however, several additional conditions under which the Wronskian vanishing in an interval implies linear dependence. For example, this is true for analytic functions. Another, probably useless, true statement is that if the Wronskian of \(n\) functions vanishes identically on an interval and the Wronksian of the first \(n-1\) of those functions never vanishes on that interval, then the \(n\)-th function is a linear combination of the others.³

Let \(A\) be a commutative ring, \(I\) an ideal of \(A\) and \(M\) an \(A\)-module. Algebraically you can define the completion \(\hat{M}\) of \(M\) as the inverse limit of the modules \(M / I^k M\) (with the canonical quotient maps \(M / I^{k+1} M \to M / I^k M\)). There is a canonical module morphism \(M \to \hat{M}\) and you can call \(M\) (\(I\)-adically) complete if this is an isomorphism.

OK, so here is the false thing I believed: the completion of an arbitrary module is complete! Rest assured that this is true if \(A\) is Noetherian. But it does fail for the simplest example of a non-Noetherian ring: take \(A = k[x_1, x_2, \ldots]\) the ring of polynomials in countably many variables, and \(M = A\). For the ideal \(I = \langle x_1, x_2, \ldots \rangle\) generated by the all variables, the completion \(\hat{M}\) is, as one would expect from the finite dimensional case, the ring of power series in countably many variables⁴. However this module is not \(I\)-adically complete: indeed, look at the sequence of polynomials \(\sum_{i=1}^n x_i^i\). If it did converge to a power series, by comparing coefficients⁵, it is clear that the limit would have to be \(\sum_{i=1}^\infty x_i^i\). But it does not in fact converge to that since it is easy to check that the tails, \(\sum_{i=j}^\infty x_i^i\) do not lie in any \(I^k \hat{M}\), i.e, you can’t have an equality of the form \(\sum_{i=j}^\infty x_i^i = m_1 g_1 + \cdots + m_l g_l\), where the \(m_i\) are finitely many monomials: every term on the RHS mentions one of the finitely many variables present in the \(m_i\), but there is no such “finite cover by variables” for the LHS.

I learned this example from Amnon Yekutieli’s paper On Flatness and Completion for Infinitely Generated Modules over Noetherian Rings.

I remember being very annoyed that I couldn’t prove that the obvious bijection \(\theta : X \wedge (Y \wedge Z) \to (X \wedge Y) \wedge Z\) is a homeomorphism. After several days I found a counterexample that I’ve since learned appears in a paper of Puppe’s:⁶ \(X = Y = \mathbb{Q}\) and \(Z = \mathbb{N}\) (all with, say, \(0\) as the basepoint).

There’s a proof by Kathleen Lewis that that is a counterexample in section 1.5 of Sigurdsson and May’s book, Parametrized Homotopy Theory. Notice that the result is stated as “\(\mathbb{Q} \wedge (\mathbb{Q} \wedge \mathbb{N})\) is not homeomorphic to \((\mathbb{Q} \wedge \mathbb{Q}) \wedge \mathbb{N}\)”, but the proof only argues that the obvious bijection \(\theta\) is not a a homeomorphism.