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.



Why do Todd classes appear in Grothendieck-Riemann-Roch formula?
You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N\_{D/X}=\mathcal O\_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O\_D$. And the Todd class pops out right away.


Indeed from the exact sequence $0\to \mathcal O\_Y(-D) \to \mathcal O\_Y\to \mathcal O\_D\to 0$, you get that 
$$ch(f\_! \mathcal O\_D)=ch(O\_Y(D))=
ch(\mathcal O\_Y) - ch(\mathcal O\_Y(-D)) = 1- e^{-D}.$$
And you need to compare this with the pushforward of $[D]$ in the Chow group, which is $D$. The ratio 
$$ \frac{D}{1-e^{-D}} = Td( \mathcal O(D) )$$ 
is what you are after. Now you have just discovered the Todd class. I suspect that this is how Grothendieck discovered his formula, too -- after seeing that this case fits with Hirzebruch's formula, that the same Todd class appears in both cases.


