Why is this map a split monomorphism?

Question

I have a question regarding a lemma in the proof of Hopkins-Neeman Correspondence. It is the beginning part of Lemma 1.2 in the The Chromatic Tower for D(R)

Let $Y$ be an object of the derived category of bounded complexes of finitely generated projectives over $R$.

The author claims there is a natural morphism, $R \to Y \otimes Y^{*}$.

I think the map he is talking about is the one corresponding to the identity under the isomorphism, $Hom(X, Y) \cong Hom(R, X^{*} \otimes Y)$.

But I am unable to understand the next claim:

If $\alpha : R \to k$ is a homorphism where $R$ is a Noetherian ring and $k$ is a field then if $ker(\alpha) \in Supp(Y)$ then

$f \otimes k : k \to Y \otimes Y^* \otimes k$ is a split monomorphism.

Can anyone help me see this?

Answer 1

Let me elaborate on what Leo says.

The morphism $f\otimes k$ lives in the derived category of bounded complexes of finite-dimensional $k$-vector spaces, which is equivalent to the category of globally finite-dimensional graded $k$-vector spaces. The equivalence is given by the cohomology functor. Since $Y$ is a bounded complex of finitely generated projective $R$-modules, we have $$H^n(Y\otimes Y^*\otimes k)=\bigoplus_{i\in\mathbb{Z}}H^{i+n}(Y\otimes k)\otimes H^{i}(Y\otimes k)^*,\qquad n\in\mathbb{Z}.$$ If $\{e_j\}_{j\in J}$ denotes a global basis of $H^{*}(Y\otimes k)$, the morphism $$f\otimes\alpha\colon k\longrightarrow\bigoplus_{i\in\mathbb{Z}}H^{i}(Y\otimes k)\otimes H^{i}(Y\otimes k)^*$$ is defined by $$1\mapsto \sum_{j\in J}e_j\otimes e_j^*.$$ Since $\dim k=1$, $f\otimes \alpha$ is either injective or trivial. Moreover, it is trivial if and only if the basis $\{e_j\}_{j\in J}$ is empty. We now use the hypothesis: this basis is non-empty becuase $H^{*}(Y\otimes k)\neq 0$ since $\ker\alpha$ is in the support of $Y$.