A technical step in the proof of Atiyah-Bott fixed point formula From John Roe, Elliptic operators, topology and asymptotic methods, page 135. 
Here Roe tried to prove that 
$$
\DeclareMathOperator{\Tr}{Tr}
\sum_{q}(-1)^{q} \Tr(Fe^{-t\bigtriangleup})=L(\zeta,\varphi)
$$ 
is independent of $t$. He differentiate it formally and verified the derivative is zero after some cancelling involving "supersymmetry". I do not really know if my "proof" is correct, so I want to ask at here. If my "proof" is correct then Roe's proof is not needed, but I believe there is something skeptical in my argument. 
I think the left hand side's value is always an integer if the individual term's value converges, because in the end it operates on $H^{*}(S_{q})$ . Since $e^{-t\bigtriangleup}$ is continuous (in the sense of its kernel) with respect to $t$, for each term involved in left hand side it must be a constant integer. And since we know the total value equals the right hand side when $t\rightarrow \infty$, they must always be the same. 
A reason this "proof" is incorrect is if we formally differentiate each term involved we get (following Roe)
$$
(-1)^{q+1}(\Tr(Fd^{*}de^{-t\bigtriangleup_{q-1}})+\Tr(F(d^{*}d)e^{-t\bigtriangleup_{q}}))
$$
and this value should not be zero as the two trace's value should not cancel.  

For readers without the book, the context of this statement is as follows: $M$ is a (compact oriented Riemannian) manifold and $\varphi:M\rightarrow M$ is a (continuous) map. Let $(S,d)$ be the Dirac complex over $M$, then it induced a map on sections:$$\varphi^{*}:C^{\infty}(S)\rightarrow C^{\infty}(\varphi^{*}S)$$where $\varphi^{*}$ acts by
$$
\varphi^{*}(f)(m)=(f(\varphi(m)),\varphi(m)),m\in M, f\in C^{\infty}(S)
$$ Suppose there is a bundle map $\zeta:\varphi^{*}S\rightarrow S$, then there is a composite map
$$
F:\zeta\varphi^{*}:C^{\infty}(S)\rightarrow C^{\infty}(S)
$$
and if $F$ is a map of complexes so that $dF=Fd$, its Lefschetz number $L(\zeta, \varphi)$ is defined by
$$
L(\zeta,\varphi)=\sum_{q}(-1)^{q}\Tr(F^{*} \textrm{on} H^{q}(S))
$$ Roe used Hodge theorem to show that
$$
\Tr(F^{*} \textrm{on} H^{q}(S))=\Tr(FP_{q}), P_{q}:L^{2}(S^{q})\rightarrow \mathcal{H}^{q}
$$
and further by using the heat kernel approximation one have
$$
\Tr(FP_{q})=\lim_{t\rightarrow \infty}\Tr(Fe^{-t\bigtriangleup})
$$
for each individual term. The question is whether the values in the individual term stays the same with $t$, or if their sum stays the same with $t$. 
 A: $\DeclareMathOperator{\Tr}{Tr}$I don't think $\Tr \, Fe^{-t\Delta_q}$ is generally an integer. (I'm not sure what you mean by "in the end it operates on $H^\ast (S_q)$".) It is only when we take the alternating sum on $q$ that we get something constant in $t$. This is the miracle of "supersymmetry".
As a simple example, take $F$ to be the identity map $C^\infty(S) \to C^\infty(S)$ (this is associated to the identity map $\varphi = \text{Id}: M \to M$, if you like). (I would think that one could come up with more interesting examples, but this will do for now.) Then $F^\ast$ is the identity on the cohomology, so $\Tr(F^\ast|_{H^q(S)} ) = \dim H^q(S)$, and the Lefschetz number is simply the Euler characteristic of the complex.
As you say, Roe's argument shows that this Euler characteristic is equal to the limit $\lim_{t\to \infty} \sum_q (-1)^q \Tr \, e^{-t\Delta_q}$, and that in fact, by "super-symmetry", $\sum_q (-1)^q \Tr \, e^{-t\Delta_q}$ is constant in $t$. (I think that this argument in this case is really just the usual McKean-Singer argument used in the heat equation proof of the index theorem.)
But the individual terms in the sum are definitely not constant in $t$. In fact, $\Tr \, e^{-t\Delta_q} \sim c_q t^{-\frac{n}{2}}$ as $t \to 0$ for some constant $c_q$, as discussed in Roe's book. So for small $t$, the terms $\Tr \, e^{-t\Delta_q}$ are quite singular, and it is somewhat miraculous that the singularities cancel in the alternating sum to give something constant in $t$. 
