# Why Hochschild cohomology is not functorial

For instance in Loday's Cyclic homology, paragraph 1.5.5 it is mentioned that *Hochschild cohomology is not functorial*. Compare this to the statement in paragraph 1.1.4 op. cit. that

*Hochschild homology is functorial*. Why this (at first sight) strange dichotomy? And why does not a single reference (of the ones I have read) addresses this issue explicitly?

Remark that this non-functoriality makes computing Hochschild cohomology of algebras a difficult task: it is not possible in general to relate it to the Hochschild cohomology of related but easier algebras.

There are two ways of showing why Hochschild cohomology is not functorial: the conceptual explanation, and a counterexample. I'll discuss both. Especially the last shows why the issue is not addressed in text books, and why this blog post perhaps should not be written, and left as an exercise. So if you are interested in the answer, try coming up with the explanation yourself first, it really is an easy exercise.

All notations are as in Loday's Cyclic homology, and familiarity with Hochschild (co)homology is assumed. Although for the counterexample it suffices to know what finite groups and their centers are.

### Conceptual

In trying to mimick the functoriality of Hochschild homology is one led to considering an algebra map $f\colon A\to B$, and $\phi\in\mathrm{Hom}_{A^{\mathrm{e}}}(\mathrm{C}^{\text{bar}}_\bullet(B),B)$. One would like to define the pullback $f^*(\phi)$ by $$ f^*(\phi)(a,a_1,\dotsc,a_n)=\phi(f(a),f(a_1),\dotsc,f(a_n)) $$ but the map $\phi$ has the wrong codomain: it should land in $A$ but it lands in $B$. The similar situation for Hochschild homology works fine though.

### Counterexample

We have the interpretation of the zeroth Hochschild cohomology as the center of the algebra. But taking the center is in general not functorial, as even the example of the maps $\mathrm{Cyc}_3\hookrightarrow\mathrm{Sym}_3\twoheadrightarrow\mathrm{Cyc}_3$ composing to the identity shows. The center in $\mathrm{Sym}_3$ is trivial, hence the composition of the centers of the morphisms is trivial too, but the center of the composition is the whole (abelian) group $\mathrm{Cyc}_3$.

To turn this into an example for Hochschild cohomology it suffices to take the group algebras on these groups, and we're done.

A similar game can by the way be played with the inclusion of the complex numbers in the quaternions.

I would like to thank Theo Raedschelders for his ideas on this (in hindsight) trivial question. The conceptual explanation is his, the mistakes in its exposition are mine.