I'm currently reading the article "An exact sequence interpretation of the Lie bracket in Hochschild cohomology" by Stefan Schwede. The thing I don't understand is the exactness in the proof of Lemma 2.1. The background is as follows:
Let $A$ be a k-algebra. Then we can form the category $$ Ext^m_A(A,A) $$ of extensions between $A$ and $A$. The maps in the category are just vertical maps between two extensions, such that the resulting diagram commutes. The statement of Lemma 2.1 is that for a given extension $$ \mathbb{E} : 0 \rightarrow A \rightarrow E_{m-1} \rightarrow ... \rightarrow E_0 \rightarrow A \rightarrow 0 $$ there is an extension $Q(\mathbb{E})$ together with a map $\varphi : Q(\mathbb{E}) \rightarrow \mathbb{E}$ such that the middle terms in $Q(\mathbb{E})$ are projective as left respectively right $A$-modules. The extension $Q(\mathbb{E})$ is constructed inductively as follows. Let $Q(E)_0$ be the free $A$-bimodule on the underlying set $E_0$. Then we have a map $Q(E)_0 \rightarrow E_0$ by the bimodule action on $E_0$. Thus we get a map $Q(E)_0 \rightarrow A$. We then construct $Q(E)_1$ as the free $A$-bimodule on the underlying set of the pullback $K_0 \times E_1$ where $K_0 = \text{Ker}(Q(E)_0 \rightarrow A)$. The map $Q(E)_1 \rightarrow Q(E)_0$ is given as the composition $$ Q(E)_1 \rightarrow K_0 \times_{E_0} E_1 \rightarrow K_0 \rightarrow Q(E)_0 $$ where the first map is again by the $A$-bimodule action.
What I don't see is how we show that the sequence $$ Q(E)_1 \rightarrow Q(E)_0 \rightarrow A \rightarrow 0 $$ is (right) exact. The article leaves this to the reader, but unfortunately I can't see how to show this.
Thank you very much for any help!
Since $Q(E)_1\to Q(E)_0\to A$ factors through $K_0\to Q(E)_0\to A$, it is clear that this composite is $0$. Likewise, it is clear that $Q(E)_0\to A$ is surjective, so let us show that $Q(E)_1\to Q(E)_0$ surjects onto $K_0$. The map $Q(E)_0\to K_0\times_{E_0}E_1$ is surjective, so it suffices to show that $K_0\times_{E_0}E_1\to K_0$ is surjective as well. The map $K_0\to E_0$ factors through $K:=\mathrm{ker}(E_0\to A)$, and the map $E_1\to E_0$ factors through $K$ too. Hence the map $K_0\times_{E_0} E_1\to K_0$ is isomorphic to the map $K_0\times_{K}E_1\to K_0$. But since $E_1\to K$ is surjective, the map $K_0\times_{K}E_1\to K_0$ must be as well by base change.