Suppose for some reason one would be expecting a formula of the kind (did I get it right?)

$$\mathop{\text{ch}}(f_!\mathcal F)\ =\ f_*(\mathop{\text{ch}}(\mathcal F)\cdot t_f)$$

valid in $H^*(Y)$ where

• $f:X\to Y$ is a proper morphism with $X$ and $Y$ smooth and quasiprojective,
• $\mathcal F\in D^b(X)$ is a bounded complex of coherent sheaves on $X$,
• $f_!: D^b(X)\to D^b(Y)$ is the derived pushforward,
• $\text{ch}:D^b(-)\to H^*(-)$ denotes the Chern character,
• and $t_f$ is some cohomology class that depends only on $f$ but not $\mathcal F$.

According to the Grothendieck–Hirzebruch–Riemann–Roch theorem this formula is true with $t_f$ being the relative Todd class of $f$, defined as the Todd class of relative tangent bundle $T_f$.

So, let's play at "guessing" the $t_f$ pretending we didn't know GHRR (I'm not sure $t_f$ is uniquely defined — you can change it by $t'$ with the property $\forall u\ f_*(\mathop{\text{ch}}(u)\cdot t') = 0$, so add conditions on $t_f$ in necessary).

Question. Expecting the formula of the above kind, how to find out that $t_f = \text{td}\, T_f$?

You don't have to show this choice works (that is, prove GHRR), but you have to show no other choice works. Also, let's not use Hirzebruch–Riemann–Roch: I'm curious exactly how and where Todd classes will appear.