Coalgebras and Coideals: Why does $\ker(\pi \otimes \mathrm{id}_C ) = I \otimes C$ hold? 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?
 A: 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 $π ⊗ \mathrm{id}_C$.
