Suppose for some reason one would be expecting a formula of the kind 

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

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](http://en.wikipedia.org/wiki/Grothendieck-Riemann-Roch%20theorem) (did I get it right?) 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 $T_f$.

So, let's play at "guessing" the $t_f$ pretending we didn't know GHRR ($t_f$ are [not uniquely defined](https://mathoverflow.net/questions/10631/homology-class-orthogonal-to-image-of-chern-characters/10660#10660), 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](http://en.wikipedia.org/wiki/Hirzebruch-Riemann-Roch_formula): I'm curious exactly how and where Todd classes will appear.