how to make the category of chain complexes into an $(\infty,1)$-category Related to this question, I would like to know, if there is an explicit presentation
of the $(\infty,1)$-category of a model category of abelian chain complexes, but this time in terms of simplicial categories. (Categories with some kind of simplicial sets as hom spaces).
The motivation is more or less the same as in the link above,i.e. as a good and understandable
example of $(\infty,1)$-categories. 
 A: I don't know what 'explicit' means, but this is all covered in Higher Algebra very well.
If you want a simplicial category at the end of the day, you can either...


*

*Take your favorite $\infty$-category (quasi-category) presentation, and straighten it to a simplicial category. (But no one wants to do this... it doesn't sound like fun.)

*Take the category of chain complexes and view at as a category enriched over itself via the internal hom. Now truncate the Hom-complexes and use Dold-Kan to get a (fibrant) simplicial set. This makes a fibrant simplicial category, which is equivalent to all the other things you might want. (For example, the homotopy category is correct, and the corresponding $\infty$-category is equivalent to one you might make from the dg-category of chain complexes, both of which agree with the $\infty$-category underlying the model category of chain complexes for a Grothendieck abelian category when this makes sense. All of your dreams come true!)


See $\S$1.3 of Higher Algebra for all the details.
A: The monoidal closed category of chain complexes was discussed in my 1964 paper Cohomology with chains as coefficients, (Proc LMS)  including the relation with the Dold-Kan functor, but not using that terminology. 
Part of the problem is that the "natural" monoidal closed structure on simplicial abelian groups, using tensor in each dimension,  is not the one that comes from chain complexes and the Dold-Kan equivalence! This problem has been recently studied by Joyal. 
In my experience  cubical methods have value in handling higher homotopies, because of the formula $I^n \times I^m \cong I^{m+n}$. There is a Dold-Kan type theorem for cubical abelian groups (with connections!). 
