In the book Rational Homotopy Theory of Félix, Halperin and Thomas they state the following. Given linear maps $f:V\rightarrow W$ and $g:W'\rightarrow V'$ between graded vector spaces then we have $\operatorname{Hom}(f,g):\operatorname{Hom}(V,V')\rightarrow\operatorname{Hom}(W,W')$ given by $$\operatorname{Hom}(f,g)(\varphi)=(-1)^{\deg f(\deg\varphi+\deg g)}g\circ\varphi\circ f.$$ Where does that sign come from? I see that it is "visually appealing" because we had to transfer $f$ through $g$ and $\varphi$ to obtain the right hand side. However, it should have an explanation in terms of the braiding isomorphism $\sigma_{V,W}:V\otimes W\rightarrow W\otimes V$ given by $\sigma_{V,W}(v\otimes w)=(-1)^{\deg v\deg w}w\otimes v$.
I am asking because I ran into trouble in a computation using the dual of a function $f^*=\operatorname{Hom}(f,\operatorname{id}_\mathbb{K})$ and the suspension convention $V[1]=\mathbb{K}[1]\otimes V$. I imagine the computation will work if I use the convention $V[1]=V\otimes \mathbb{K}[1]$, but would like to know the origin of my problems.
As a side question, I have yet to find a nice soft introduction to these subtleties in graded algebra that are suitable for a beginner in algebra without much knowledge of category theory. Any suggestions?
Ok, I found a partial answer. First of all, note that the isomorphim $V\rightarrow V^{**}:v\mapsto\bar{v}$ has a sign. Namely, $\bar{v}(f)=(-1)^{\deg f\deg v}f(v)$. For this I have no good explanation though so if anyone has an idea I would appreciate it.
In any case, if one is willing to accept that, consider the chain of isomorphisms $$\operatorname{Hom}(V,W)\rightarrow W\otimes V^*\rightarrow W^{**}\otimes V^*\rightarrow V^*\otimes W^{**}\rightarrow\operatorname{Hom}(W^*,V^*)$$ Then the second and third arrows have koszul signs. If $a_i$ is a basis of $V$, $b_i$ of $W$, and we denote the corresponding dual basis by upper indices, under this chain of maps we have $$f\mapsto b_i\otimes f^i_j a^j\mapsto\bar{b_i}\otimes f^i_ja^j\mapsto (-1)^{\deg b_i\deg a_j}f^i_ja^j\otimes\bar{b_i}\mapsto f^*, $$ where $$f^*(b^k)=(-1)^{\deg b_i\deg a_j+\deg b_i\deg b_k}f^i_j a^j\delta^k_i=(-1)^{\deg b_k(\deg a_j+1)}f^k_j a^j.$$ We then have $$f^*(b^k)(a_i)=(-1)^{\deg b_k(\deg a_i+1)}f^k_i=(-1)^{\deg b_k(\deg a_i+1)}b^k(f(a_i)).$$ Noting that if $f^k_i\neq 0$ we then have $\deg b_k=\deg f+\deg a_i$ we conclude that $$f^*(b^k)=(-1)^{\deg b_k(\deg b_k+\deg f+1)}b^k\circ f=(-1)^{\deg f\deg b_k}b^k\circ f.$$ This explains the sign I was concerned with primarily. However, where does the identification of $V$ and $V^{**}$ that I used come from?