Universal coefficients theorem for a field Let $F$ be a field and $C$ be a chain complex (of free abelian groups).  I am confused about what the isomorphism $\phi : H^p(C; F) \to \hom_F(H_p(C;F),F)$ is exactly.
I know that there is a natural map $H^p(C; F) \to \hom_\mathbb{Z}(H_p(C;\mathbb{Z}),F)$ (the "Kronecker map") which is the one appearing in the universal coefficients theorem.  I have also seen an isomorphism of chain complexes of vector spaces
\begin{align*}
\tau: \hom_\mathbb{Z}(C_p, F) &\to \hom_F(C_p \otimes F, F) \\
 f         &\mapsto c \otimes \alpha \mapsto \alpha f(c)
\end{align*}
where this first chain complex is the chain complex for $H^p(C; F)$.  
Is the universal coefficients isomorphism $\phi$ given by $\phi([f])([c]) = \tau(f)(c)$?  
 A: Here are the details. For all $p$, you have a natural map: 
$$\tau^p:\operatorname{Hom}_{\mathbb{Z}}(C_p,F)\to\operatorname{Hom}_{F}(C_p\otimes F,F),\quad (\tau^p(f))(x\otimes a) = af(x)$$
(the tensor-hom adjunction in disguise) which, when applied to the differential $d_p:C_p\to C_{p-1}$ gives rise to a commutative square:
$$
\require{AMScd} \begin{CD}
    \operatorname{Hom}_{\mathbb{Z}}(C_{p-1},F) @>{d^p}>> \operatorname{Hom}_{\mathbb{Z}}(C_{p-1},F)\\ @V{\tau^{p-1}}VV @VV{\tau^p}V\\
    (C_{p-1}\otimes F)^* @>>{(d_p\otimes 1)^*}> (C_{p-1}\otimes F)^*
    \end{CD}
$$
where $(C_p\otimes F)^*=\operatorname{Hom}_{F}(C_p\otimes F,F)$ is the usual dual space notation. That is, the $\tau^p$ constitute a cochain map between the complexes $\operatorname{Hom}_{\mathbb{Z}}(C,F)$ and $(C\otimes F)^*$. Since the $\tau^p$ are isomorphisms, the induced map on cohomology:
$$\tau^*:H^p(C,F)\to H^p((C\otimes F)^*),\quad \tau^*[f] = [\tau^p(f)]$$
is an isomorphism. However, since $F$ is an injective $F$-module, the (contravariant) functor $\operatorname{Hom}_F(-,F)$ is exact, and therefore we have an isomorphism
$$H^p(\operatorname{Hom}_F(C\otimes F,F)) \cong \operatorname{Hom}_F(H_p(C;F),F)$$
which sends a an element $[g:C_p\otimes F\to F]$ to $\bar{g}:H_n(C;F)\to F$, where $\bar{g}[x\otimes a] = g(x\otimes a)$. Composing these two isomorphisms gives:
$$[f]\mapsto [\tau^p(f)]\mapsto \overline{\tau^p(f)},$$
where $\overline{\tau^p(f)}[x\otimes a]=\tau^p(f)(x\otimes a)$. So the map does act as you expected.
The isomorphism between $H^n(C;F)$ and $H_n(C;F)^*$ follows from similar reasoning found here.
