# Do direct limits (filtered colimits) commute with pullbacks, in C*-algebras?

I asked this question already on math.stackexchange, but maybe it is also useful to ask this here, since it was not answered there.

Suppose we have three directed sequences of $C^*$-algebras, say $(A_n,\varphi_n)$,$(B_n,\psi_n)$ and $(C_n,\theta_n)$ and $*$-homomorphisms $\alpha_n:A_n\rightarrow C_n$ and $\beta_n:B_n\rightarrow C_n$, then we can take the pullback $A_n\times_{C_n}B_n$ for all $n\in\mathbb{N}$ and can also take the direct limit, thus $\lim_{n\rightarrow\infty}{A_n\times_{C_n}B_n}$. My question is: Does the following hold: $$\lim_{\rightarrow}{A_n\times_{C_n}B_n}=\lim_{\rightarrow}{A_n}\times_{\lim_{\rightarrow}{C_n}}\lim_{\rightarrow}{B_n}$$ or in other words: do direct limits preserve pullbacks? From my point of view this does not hold in general but I can not find a good argument. Someone an idea? Thank you very much.

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Both at math.se and here, the title was rather unclear, which I think is why there’s been some misunderstanding and negative reactions in comments (including thinking that the question was more trivial than it is). I’ve edited the title to give the actual question a bit more clearly; hopefully I’ve understood OP’s intentions correctly. – Peter LeFanu Lumsdaine Jul 29 '14 at 13:28
Please do not rush to close this question unless you are sure it is trivial. I somehow believe $C^\ast$-algebras are locally $\aleph_1$-presentable, but not locally $\aleph_0$-presentable, and this makes a big difference to the question. (If it were locally $\aleph_0$-presentable, then certainly filtered colimits would commute with finite limits.) – Todd Trimble Jul 29 '14 at 13:48
To find a counterexample, we could start by concentrating on commutative C*-algebras. The full subcategory Comm-C*-Alg of C*-Alg is closed under limits and colimits, so it doesn't matter which category we think of the (co)limits as happening in. Now Comm-C*-Alg is dual to the category CptHff of compact Hausdorff spaces, so to find a counterexample, it's enough to show that inverse lims (cofiltered lims) don't commute with pushouts in CptHff. Taking inverse lims does preserve coproducts and epis in CptHff, so those two simple kinds of pushout won't help us... we need something a bit cleverer. – Tom Leinster Jul 29 '14 at 15:13
@TomLeinster I like the general thrust of your comment, but I don't understand why the subcategory of commutative $C^\ast$-algebras is closed under colimits, in particular under coproducts. I would have thought this failed for the same reason that commutative rings are not closed under coproducts taken in the category of all rings. – Todd Trimble Jul 29 '14 at 16:23
@TomLeinster On second thought, I think you're okay: just replace "colimit" by "filtered colimit", and I think your statement is correct (and that's the only kind of colimit we need consider). – Todd Trimble Jul 29 '14 at 18:17

I may be showing my ignorance of category theory, but don't think this is true. Work on $l^2$. Let $\mathcal{A}_n$ consist of the operators $A$ satisfying $\langle Ae_i, e_j\rangle = 0$ if $\max(i,j) > n$ (so $\mathcal{A}_n$ is isomorphic to the $n\times n$ matrices). The direct limit of the $\mathcal{A}_n$ is the compact operators on $l^2$. Let $P$ be a rank one projection that is not contained in any of the $\mathcal{A}_n$, let $\mathcal{B} = \mathbb{C}\cdot P$ be the C*-algebra it generates, and let $\mathcal{B}_n = \mathcal{B}$ for all $n$. Also let $\mathcal{C}_n = B(l^2)$ for all $n$. All the morphisms are inclusion maps.
Each pullback $\mathcal{A}_n\times_{\mathcal{C}_n} \mathcal{B}_n$ is zero, so the direct limit of the pullbacks is zero, but the pullback of the direct limits is (isomorphic to) $\mathcal{B}$.
Or better yet, define $\mathcal{B}_n$ the same way as $\mathcal{A}_n$, but with respect to a different basis so that $\mathcal{A}_n \cap \mathcal{B}_n = \{0\}$ for all $n$. Then the direct limit of the pullbacks is zero, but the pullback of the direct limits is $K(l^2)$. – Nik Weaver Jul 29 '14 at 18:47