When existence of transfer map of algebraic $K$-theory implies rational injection. Given a map of smooth projective varieties $f:X\rightarrow Y$ over fields, there is a projection formula in algebraic $K$-theory given by $f_*(\alpha.f^*(\beta))=\beta.f_*(\alpha)$. I was wondering whether this implies that algebraic $K$-group of $Y$ rationally injects into $X$?
The natural approach is to plug in $\alpha$ to be the class of trivial line bundle $[\mathcal{O}_X]$ in $K_0(X)$. This implies that $f_*f^*\beta=\beta.f_*([\mathcal{O}_X])$. So the problem is identifying $f_*([\mathcal{O}_X])$ and seeing whether multiplication by it is rationally an isomorphism or not. This leads to GRR and some complicated formulas which lead me to think this is not true in general. Is there a situation that it becomes true (other than the famous etale case)? Does the transfer induce rational injection if the map is finite?
This is I think true for finite maps. If the generic degree is $d$ then it seems for any line bundle $L$, according to some intersect theoretic facts, $f_*L$ has chern classes $d$ times image of chern classes under $f$ (as cycles). So for trivial line bundle it is not going to have any chern classes and its rank is going to be $d$. So multiplication by $f_*([\mathcal{O}_X])$ will be like multiplying with $d$ rationally.
I'm not sure whether this is true for ramified covers.
 A: I believe the following holds: If $X/\mathbb{Z}$ is of finite type and $X_t:= X\times_{\mathbb{Z}} \mathbb{F}_p$
is the fiber of $X$ at $t:=(p)\neq (0)$ it follows $K_m(X_t)_{\mathbb{Q}}=0$ is a conjecture, the "Parshin conjecture":
https://en.wikipedia.org/wiki/Parshin%27s_conjecture
There is the canonical pull back map
$$f^*: K_m(X)_{\mathbb{Q}} \rightarrow K_m(X_t)_{\mathbb{Q}}.$$
The group $K_m(X)_{\mathbb{Q}}$ is non-zero in general.
Question: "Given a map of projective varieties $f:X→Y$, there is a projection formula in algebraic K-theory given by $f∗(α.f∗(β))=β.f∗(α)$. I was wondering whether this implies that algebraic K-group of $Y$ rationally injects into $X$?"
Answer: Take $X:=\mathbb{P}^n_{\mathbb{Z}}$. You get the map
$$f^*: K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \rightarrow K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}$$
and the left side is non-zero, the right side is zero by the projective bundle formula:
$$K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}\cong K_*(\mathbb{F}_p)_{\mathbb{Q}}\otimes \mathbb{Q}[t]/(t^{n+1})=0$$
since $K_*(\mathbb{F}_p)_{\mathbb{Q}}=0$. It follows $K_*(\mathbb{P}^n_{\mathbb{F}_p})_{\mathbb{Q}}=0$.
We get
$$K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \cong K_*(\mathbb{Z})_{\mathbb{Q}}\otimes \mathbb{Q}[t]/(t^{n+1})$$
and $K_*(\mathbb{Z})_{\mathbb{Q}} \neq 0$ hence $K_*(\mathbb{P}^n_{\mathbb{Z}})_{\mathbb{Q}} \neq 0$.
Hence $f^*$ is not injective in general as asked in the question.
Example: (The "projective bundle formula") The projection map $\pi: X:=\mathbb{P}(E^*) \rightarrow S$
where $S \subseteq \mathbb{P}^n_k$ is a smooth projective variety over $k$,
is smooth and there is - again by the projective bundle formula - an injective map
$$ \pi^*: K_*(S)_{\mathbb{Q}} \rightarrow K_*(X)_{\mathbb{Q}}$$
but the pull back map is not an injection in general.
Example: (The disjoint union). Let $Y:= \cup_i X_i$ where $X_i \cong X$ for all $i$, with $X$ smooth and where the union is the disjoint union. You get a "canonical" injection map $j: X \rightarrow Y$, and the pull back map $j^*$ is the map
$$j^*: K_*(Y)_K \cong \oplus_i K_*(X)_K \rightarrow K_*(X)_K .$$
The map $j^*$ is the projection onto one of the direct summands, and hence $j^*$ is not an injection (here $K$ is any field).
