TITLE: Why do Todd classes appear in Grothendieck-Riemann-Roch formula?
QUESTION [57 upvotes]: 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 (did I get it right?) 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 ($t_f$ are not uniquely defined, 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.

REPLY [3 votes]: A conceptual answer to "exactly how and where Todd classes will appear:"
You don't have to use Todd classes if you stick to Hochschild homology. This might not be the best reference, but I learned this from 1.2.1 in
https://arxiv.org/pdf/1804.00879.pdf
You can stick to just the Chern character and factor through HH(Perf(X)); the Todd class corrects the fact that the HKR isomorphisms $HH(Perf(X)) \simeq H^*(X, \bigoplus \Omega_X^p)$ are NOT natural in $X$ until correction by Todd.
My apologies -- I'm sure there's a more down-to-earth reference of which I'm not aware.