Consider a representation $\,A(g)\,$ of a group $\,G\,$ in a vector space $\,\mathbb{V}\,$. $~$For a subspace $\,{\mathbb{V}}_1\subseteq{\mathbb{V}}\,$ invariant under $\,A\,$, $$ A(g)~{\mathbb{V}}_1\,\subseteq\,{\mathbb{V}}_1~~, $$ ${\mathbb{V}}\,$ can be split into classes, so that two vectors belong to the same class if their difference lies in $\,{\mathbb{V}}_1\,$. $~$By choosing a class representative $\,v\,$, we can present a class as $\,v + {\mathbb{V}}_1\,$, $\,v\in{\mathbb{V}}\,$. These classes constitute the factor space $\,{\mathbb{V}}/{\mathbb{V}}_1\,$ where we define the factor representation $$ (A/A_1)(g)\,(v\,+\,{\mathbb{V}}_1)\;\equiv\;A(g)\,v\;+\;{\mathbb{V}}_1\;\;.\qquad\quad(1) $$ The canonical projection $\,\pi\,$ maps a vector to its class, $$ \pi\;v\;=\;v\;+\;{\mathbb{V}}_1\;\;, $$ and intertwines $\,A\,$ with $\,(A/A_1)\,$: $$ (A/A_1)(g)\;\pi\, v \;=\; (A/A_1)(g)\;(v\,+\,{\mathbb{V}}_1) \;=\; A(g)\,v\;+\;{\mathbb{V}}_1 \;=\; \pi\,A(g)\, v\,\;. $$
Let the representations $\,A(g)\,$ and $\,A^{\,\prime}(g)\,$ in spaces $\,\mathbb{V}\,$ and $\,\mathbb{V}^{\,\prime}(g)\,$ be intertwined by a morphism $\,M\;$: $$ M~A(g)~=~A^{\,\prime}(g)~M~~. $$ For brevity, I shall call the kernel $\,\mbox{Ker}\,M\,$ simply as $\,$Ker, the image $\,\mbox{Im}\,M\,$ as $\,$Im$\,$. Both are invariant subspaces. Each supports a subrepresentation: $$ B(g)\,v~\equiv~A(g)\,v\Big{|}_{v\in\,\rm{Ker}}\quad,\qquad B^{\,\prime}(g)\,v^{\,\prime}~\equiv~A^{\,\prime}(g)\,v^{\,\prime}\Big{|}_{v^{\,\prime}\in\,\rm{Im}}\;\;. $$
$$ ~~ $$ $\quad$ QUESTION: $~~$is it true that $\,(A/B)\,\simeq\,B^{\,\prime}~$? $$ ~~ $$ In the case of $\,{\mathbb{V}}_1\,=\;\mbox{Ker}\,$, equation (1) becomes: $$ (A/A_1)(g)\,(v\,+~\mbox{Ker})\;\equiv\;A(g)\,v\;+\;{\mbox{Ker}}\;\;.\qquad\quad(2) $$ While the morphism $\,M\,$ is defined as an operator acting on vectors in $\,{\mathbb{V}}\,$, its definition can be extended, to consider an action of this operator on a whole class: $$ M\;(A/A_1)(g)\,(v\,+~\mbox{Ker})\;\equiv\;M\;A(g)\,v\;+\;M\;{\mbox{Ker}} $$ $$ \;=\;A^{\,\prime}(g)~M~v~+~\vec{\,0}^{\,\prime} \;=\;B^{\,\prime}(g)~M~v \;\;.\qquad\quad(3) $$ We see that $\,M\,$ intertwines $\,(A/A_1)\,$ and $\,B^{\,\prime}\,$. However, my understanding is that this does not necessitate $\,(A/B)\,\simeq\,B^{\,\prime}\,$ unless we impose an additional requirement that $\,A\,$ is irreducible (and $\,\mbox{Ker}\,=\,\vec{\,0}\,$).
Or am I missing something? Say, can we agree to define $\,M\,$ not on vectors but on classes of vectors, and thereby make $\,M\,$ always invertible?