# Why is Hochschild cohomology $HH^n(A, A)$ not a functor of $A$?

Let $$k$$ be a commutative ring, $$A$$ a $$k$$-algebra and $$HH^n(A, M)$$ the $$n$$-th Hochschild cohomology of $$A$$ with coefficients in the $$A$$-bimodule $$M$$. In the book Cyclic Homology by Loday the following curious few lines appear on page 40 when talking about the form that the coefficients can take:

For $$M=A$$ the groups $$HH^n(A, A)$$ have been extensively studied in the literature because they are related to deformation theory. But one should note that they are not functors of $$A$$. However if $$M = A^* = \mbox{Hom}_k(A, k)$$, then the groups $$HH^n(A, A^*)$$ are indeed functors of $$A$$.

Now I'm quite confused by this assertion that $$HH^n(A, A)$$ isn't a functor of $$A$$, because I know that you can interpret the $$n$$-th cohomology as the derived functor

$$HH^n(A, M) = \mbox{Ext}_{A^e}^n(A, M)$$

where $$A^e = A\otimes_k A^{op}$$ is the enveloping algebra of $$A$$. In fact, it's mentioned on the very next page of the book. So clearly each $$HH^n(A, M)$$ is a functor if I've understood this correctly. So presumably the important part is that $$HH^n(A, A)$$ is not a functor "of $$A$$", but I'm not entirely sure what this means. I'd be grateful if someone could help me understand how it is that $$HH^n(A, A)$$ isn't a functor of $$A$$.

Edit:

Thanks to the comments beneath the question (thanks for your help guys, I appreciate it) I think I have a better idea of what's going on. So using the example of $$\mbox{Hom}_k(A, A)$$, the reason why this isn't a functor of $$A$$ is because given another $$k$$-algebra $$B$$, and a map $$f:A\to B$$, there is no real way of assigning $$f$$ to a map

$$\mbox{Hom}_k(A, A)\to\mbox{Hom}_k(B, B)$$

However, now that I understand this, it seems to me $$\mbox{Hom}_k(A, A^*)$$ suffers from the same issue. So how is it that (as mentioned in the paragraph I cite above) $$HH^n(A, A^*)$$ is a functor of $$A$$? Am I just missing some obvious way of assigning $$f:A\to B$$ to $$\mbox{Hom}_k(A, A^*)\to\mbox{Hom}_k(B, B^*)$$?

• $A$ appears in the formula $\textrm{Ext}^n_{A^e} (A, A)$ several times, but not all with the same variance. So it isn't obviously a functor of $A$. (Recall that a functor acts on objects and morphisms. Can you define the action on morphisms here?) Aug 4 '20 at 11:16
• Well, for example, $\textrm{Hom} (X, Y)$ is contravariant in $X$ and covariant in $Y$. Thus $X$ appears in the expression $\textrm{Hom} (X, X)$ with mixed variance – and this is essentially the reason why $\textrm{End} (X)$ is not a functor of $X$. Hochschild cohomology is just a souped up version of this. Aug 4 '20 at 12:05
• @SeraPhim Maybe it's easier to see in the case of $\mathrm{Hom}(X,X)$ in some category, which is basically the same issue. If you want to make this a functor, then given $f:X\to Y$ you have to define a map in one direction or the other between $\mathrm{Hom}(X,X)$ and $\mathrm{Hom}(Y,Y)$, and in general there is no reasonable way to do this. Aug 4 '20 at 16:50
• @SeraPhim: Given a map $f \colon X \rightarrow Y$, you have an induced pullback map $f^{*} \colon \operatorname{Hom}_k(Y,Y^{*}) \rightarrow \operatorname{Hom}_k(X,X^{*})$ given by $((f^{*}(\phi))(x))(x') = (\phi(f(x)))(f(x'))$. Similarly, you can define a map $f^{*} \colon C^n(B,B^{*}) \rightarrow C^n(A,A^{*})$ which commutes with the differential and induces a map on the Hochschild cohomology. Aug 6 '20 at 11:36
• The point is that dualisation $(-)^*$ is a contravariant functor. Thus $\textrm{Hom} (X, X^*)$ is purely contravariant in $X$. More to the point, functors can be composed, and we have functors $X \mapsto (X, X^*)$ and $(X, Y) \mapsto \textrm{Hom} (X, Y)$, so we have the composite functor $X \mapsto \textrm{Hom} (X, X^*)$. By contrast $X \mapsto (X, X)$ and $(X, Y) \mapsto \textrm{Hom} (X, Y)$ cannot be composed because of incompatible variance. Aug 6 '20 at 11:59

$$A$$ appears in the formula $$\textrm{Ext}^n_{A^e} (A, A)$$ several times, but not all with the same variance. So it isn't obviously a functor of $$A$$.
[...] $$\textrm{Hom} (X, Y)$$ is contravariant in $$X$$ and covariant in $$Y$$. Thus $$X$$ appears in the expression $$\textrm{Hom} (X, X)$$ with mixed variance – and this is essentially the reason why $$\textrm{End} (X)$$ is not a functor of $$X$$. Hochschild cohomology is just a souped up version of this.
Maybe it's easier to see in the case of $$\mathrm{Hom}(X,X)$$ in some category, which is basically the same issue. If you want to make this a functor, then given $$f:X\to Y$$ you have to define a map in one direction or the other between $$\mathrm{Hom}(X,X)$$ and $$\mathrm{Hom}(Y,Y)$$, and in general there is no reasonable way to do this.