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 $Y$ 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](http://en.wikipedia.org/wiki/Grothendieck-Riemann-Roch%20theorem) this formula is true with $t_f$ being the relative Todd class of $f$, defined as the [Todd class](http://en.wikipedia.org/wiki/Todd_class) of relative tangent bundle $T_f$.

Now, I want to somewhow "derive" the above formula, but I suspect $t_f$ is not uniquely defined; perhaps you can change it by a homology class $t'$ with the property $\forall u\ f_*(\mathop{\text{ch}}(u)\cdot t') = 0$. 

> **Question.** What extra conditions we must put on $t_f$  to make sure the only possibility is $t_f = \text{td}\, T_f$?

I think one could add some kind of functoriality for $t_f$. You don't have to show this choice works, but you have to show no other choice works.
Also, let's not use [Hirzebruch–Riemann–Roch](http://en.wikipedia.org/wiki/Hirzebruch-Riemann-Roch_formula): I'm curious exactly how and where Todd classes will appear.