It would be good to draw a diagram here, but as far as I understand, it's impossible. This resembles the notion of fiber sum, so I hope it is well-known.

Let $A$ and $B$ be (unital) algebras over $\mathbb C$, $X$ a bimodule over $A$,
$\varphi:A\to B$ a homomorphism of algebras, and $\alpha:A\to X$ a morphism of $A$-bimodules. Let us call a triple $(Y,\beta,\psi)$ an $A$-$B$-fiber sum for $(A,B,M,\varphi,\alpha)$, if

  1. $Y$ is a $B$-bimodule,

  2. $\beta:B\to Y$ is a morphism of $B$-bimodules,

  3. $\psi:X\to Y$ is a morphism of $A$-bimodules,

  4. $\beta\circ\varphi=\psi\circ\alpha$,

  5. if $(Y',\beta',\psi')$ is another triple with the properties 1-4, then there exists a morphism $\upsilon:Y\to Y'$ such that the diagram that I hope everybody understands, is commutative.

So the question:

Does the $A$-$B$-fiber sum $(Y,\beta,\psi)$ exist for each $(A,B,M,\varphi,\alpha)$?

  • $\begingroup$ I'm guessing you want some universal property here, like $Y$ being initial with this property ? Otherwise you can take the zero bimodule... $\endgroup$ Commented Mar 13, 2016 at 16:20
  • $\begingroup$ Ah, yes! Like in the definition of the fiber sum. I'll edit this. $\endgroup$ Commented Mar 13, 2016 at 16:26

1 Answer 1


As you say, using commutative diagrams can be a little dangerous since we work simultaneously in different categories, so I will give an elementary solution (pun intended : it means "that uses elements instead of arrows").

A $A$-bimodule morphism $A\to X$ is the same thing as the choice of an element $x_0\in X$ (given by the image of $1$) satisfying $ax_0 = x_0a$ for all $a\in A$. Likewise, a $B$-bimodule morphism $B\to Y$ is the choice of a $y_0\in Y$ such that $by_0=y_0b$ for all $b\in B$.

So you are given a $x_0\in X$ that has this property, and you are looking for a morphism $\psi: X\to Y$ such that $y_0:= \psi(x_0)$ has this property in $Y$.

First, to give a $A$-bimodule morphism $\psi: X\to Y$ where $Y$ is a $B$-bimodule is the same as giving a $B$-bimodule morphism $\widetilde{\psi}: \widetilde{X}\to Y$ where $\widetilde{X} = B\otimes_A X\otimes_A B$. And if you put $\widetilde{x_0} = 1\otimes x_0\otimes 1$, you want $y_0:= \widetilde{\psi}(\widetilde{x_0})$ to satisfy the commutation property above.

Now the universal (in fact initial) choice is to take the quotient map $\widetilde{X}\to \widetilde{X}/I$ Where $I$ is the sub-$B$-bimodule generated by the $b\widetilde{x_0}-\widetilde{x_0}b$ for all $b\in B$.

It should give the answer you're looking for.

  • $\begingroup$ I think I'm toying with the Hochschild cohomology groups $HH^0(A,X)$ and $HH^0(B,Y)$ here, but I'm don't know much about this cohomology, so I don't mention it. $\endgroup$ Commented Mar 13, 2016 at 16:44
  • $\begingroup$ Thank you! $\phantom{.}\kern-15pt$ $\endgroup$ Commented Mar 13, 2016 at 16:54

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