I recently went to a talk where the speaker mentioned that
$$ 0 \rightarrow SU(n) \rightarrow SU(n+1) \rightarrow S^{2n + 1} \rightarrow 0$$
was exact. I think it's well-known that this sequence is a fibration, but is it an exact sequence too?
For this to the the case we would need to put a group structure on $S^{2n + 1}$ and define a group homomorphism from $SU(n+1)$ to it... but I don't know how to do this. There is a bijection between $S^{2n+1}$ and cosets $SU(n+1)/SU(n)$, but since $SU(n)$ is not normal in $SU(n+1)$, $SU(n+1)/SU(n)$ does not have a group structure. Hence my source of confusion.
Was the speaker correct, or did they mean something else by 'exact'?
This is not a standard usage of "exact", and looks like a misstatement. In particular, $S^{2n+1}$ does not admit any topological group structure if $n>1$. I imagine they really just meant that it's a fiber sequence.