Is $\mathbb{C}[T \times_T T] \cong \mathbb{C}[T]$?

Question

Let $T$ be an algebraic group. Do we have the following result: The algebra $\mathbb{C}[T \times_T T]$ is equal to the image of the co-multiplication $\mathbb{C}[T] \to \mathbb{C}[T] \otimes \mathbb{C}[T]$? Here the co-multiplication $\mathbb{C}[T] \to \mathbb{C}[T] \otimes \mathbb{C}[T]$ is the pull-back of the multiplication map $T \times T \to T$ and $T \times_T T$ is the fiber product over $T$. Is $\mathbb{C}[T \times_T T] \cong \mathbb{C}[T]$? Thank you very much.

Answer 1

In any category $\mathsf{C}$, if $X$ and $Y$ are objects with an arrow $f : Y \to X$, the fibered product $X \times_X Y$ is always isomorphic to $Y$ (pretty much directly from the definitions). In your case you get $T \times_T T \cong T$, hence $\mathbb{C}[T \times_T T] \cong \mathbb{C}[T]$.

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If you want to convince yourself that $X \times_X Y \cong Y$, you can prove it as such: $$\begin{align} \hom(Z, X \times_X Y) & \cong \hom(Z, X) \times_{\hom(Z, X)} \hom(Z, Y) \\ & \cong \hom(Z, Y). \end{align}$$ The second bijection is $(u,v) \mapsto v$ (where $u : Z \to X$ and $v : Z \to Y$). This is a bijection because, since you're in the fibered product, you necessarily have $u = f \circ v$. Hence by the Yoneda lemma $X \times_X Y \cong Y$.