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In a proof on comodules and coideals I found the following passage:

Let $C$ be a coalgebra, and $I \subset C$ a vector subspace.

Let $\pi \colon C \rightarrow C/I $ be the canonical projection. Consider the linear map $f := (\pi \otimes \mathrm{id}_C) \circ \Delta \colon C \rightarrow C/I \otimes C$.

By the universal property of quotient vector spaces, there is a unique map $F \colon C/I \rightarrow C/I \otimes C$ with $F \circ \pi = f$ if and only if $I \subset \ker(f)$. Now $I \subset \ker(f)$ is equivalent to $\Delta(I) \subset \ker(\pi \otimes \mathrm{id}_C) = I \otimes C$, i.e., $I$ is a right coideal.

Why does the last equality $\ker(\pi \otimes \mathrm{id}_C ) = I\otimes C$ hold?

Jendrik Stelzner
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Max Demirdilek
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    Well, $\ker(\pi) = I$ and $\ker(\operatorname{id}_C) = {0}$, so the kernel of $\pi \otimes id_C$ is $I \otimes C + C \otimes {0} = I \otimes C$. – Malkoun Mar 13 '20 at 08:56
  • Thanks a lot, are you using https://math.stackexchange.com/a/541554/752637? – Max Demirdilek Mar 13 '20 at 09:26
  • I kind of assumed it was true. I should be more careful. That the sum in my previous comment is a subset of the kernel of the tensor product map is clear. But the other inclusion is less clear in general. I think this is what they discuss in the answer you linked to. – Malkoun Mar 13 '20 at 09:49

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For every two $$-linear maps $$ f_1 \colon V_1 \to W_1 \,, \quad f_2 \colon V_2 \to W_2 $$ between vector spaces, we have the equality $$ \ker(f_1 ⊗ f_2) = \ker(f_1) ⊗ V_2 + V_1 ⊗ \ker(f_2) \,. $$ (If $f_1$ and $f_2$ are both surjective, i.e., quotient homomorphisms, then this is also true if we replace $$ by an arbitrary commutative ring.) Proofs of this can be find in the question What is the kernel of the tensor product of two maps?. It follows in the given situation that $$ \ker(π ⊗ \mathrm{id}_C) = \ker(π) ⊗ C + C ⊗ \ker(\mathrm{id}_C) = I ⊗ C + C ⊗ 0 = I ⊗ C \,. $$

Alternatively, we could consider the short exact sequence $$ 0 \longrightarrow I \longrightarrow C \xrightarrow{\enspace π \enspace} C/I \longrightarrow 0 \,. $$ Tensoring over $$ is exact because it is a field, whence the sequence $$ 0 \longrightarrow I ⊗ C \longrightarrow C ⊗ C \xrightarrow{\enspace π ⊗ \mathrm{id}_C \enspace} (C/I) ⊗ C \to 0 $$ is again exact. This tells us that $I ⊗ C$ is the kernel of $π ⊗ \mathrm{id}_C$.

Jendrik Stelzner
  • 17,081