Is the continuous dual of a topological chain complex chain equivalent to the algebraic dual? I apologize in advance if this is a naive question. 
Def: A topological chain complex is a chain complex of topological $\mathbb{R}$-vector spaces such that the boundary maps are continuous.
Let $C$ be a topological chain complex.
one can in a natural way consider the dual $C$* of the chain complex $C$ (ignoring the topology).
One can also consider the continuous dual of the topological chain complex $C^{*c}$  (because the boundary maps of $C$ are continuous)
Are $C$*, $C^{*c}$ chain equivalent? I would also like to know if the answer is yes after possibly adding some more conditions to ensure nothing pathological happens. For example, is the answer yes if one assumes further that all homologies of C are finite dimensional ?
Thank you, 
 A: In general they are not chain equivalent, even if $H_\ast(C)$ is finite dimensional. (I need the axiom of choice) 
For a counterexample let $p \in (0,1)$ and let $f: L^p([0,1]) \to \mathbb{R}$ be a non-zero linear functional. $W := \ker(f)$ is a topological space with the subspace topology and the inclusion $W \hookrightarrow L^p([0,1])$ is continuous. Take
$$C = (C_2 \to C_1 \to C_0):\qquad W \hookrightarrow L^p([0,1]) \to 0$$
Then $H_0(C)=H_2(C)=0, H_1(C)=\mathbb{R}$. In particular, $H_\ast(C)$ is finite dimensional. 
By the Universal Coefficient Theorem, $H_1(C^\ast) \cong H_1(C)^\ast=\mathbb{R}$. 
Denote the continuous dual by $(\,\, )'$. Then 
$$C': \qquad 0 \to L^p([0,1])' \to W' $$
Since $L^p([0,1])' = 0$ (https://en.wikipedia.org/wiki/Lp_space), the complex is 
$$C': \qquad 0 \to 0 \to W'$$
and hence $H_1(C')=0 \not\cong H_1(C^\ast)$. Since chain equivalent complexes have isomorphic homology groups, $C^\ast$ and $C'$ are not chain equivalent. 
Remark: One might ask, what $W'$ is. If I'm not mistaken, one can adopt the proof of the BLT theorem (https://en.wikipedia.org/wiki/Continuous_linear_extension) to the metric of $L^p([0,1])$ and show that each continuous linear functional $W \to \mathbb{R}$ extends uniquely to a continuous linear functional $L^p([0,1])\to \mathbb{R}$. This yields $W'=0$.  
