For a short exact sequence $1 \to A \stackrel{\iota}{\to} B \stackrel{\pi}{\to} C \to 1$ we have the situation that the square

is both a pushout in groups and also a pullback.

This would be wrong if $A$ is an arbitrary subgroup of $B$, but is true because it is a normal subgroup.

For a field $k$ I can also take the group algebras and consider the square

in differential graded $k$-algebras (all concentrated in degree zero).

This is still a pullback, and since $\pi$ is a fibration and all objects are fibrant, it is also a homotopy pullback.

However, according to my computation it does not seem to be a homotopy pushout (i.e. we have no stability property of the square): If I compute the homotopy pushout of

then I get something which is in a sense "bigger" than $k[C]$.

Is there a way how this can be fixed or a good explanation for that? Can I write down a different homotopy pushout (or more generally a homotopy colimit) that allows me to recover $k[C]$ up to equivalence?

Note: This question is related to Homotopy quotient of groups

Thanks for any hints.