Clarification about $C \otimes D$ with $C,D$ contractible chain complex The question is strictly related to Eilenberg-Zielber theorem, since the ones I'm about to write are the Propositions needed in order to prove the Theorem.
Given $(C,d_c),(D,d_d)$ chain complex and defined $C \otimes D$ as usual, I have the following lemma :

Lemma : Given a product of complex such that one of two is
contractible (i.e identity chain homotopic to $0$) $C \otimes D$ is
also contractible.

I do understand the proof of the Lemma what I don't get is : the complex here are supposed to be reduced chain complex (i.e $\tilde{C}$) ?
This question arises from the corollary given to me to this theorem :

Corollary :
$1)$ If $X$ is contractible then $\tilde{C}(X) \otimes \tilde{C}(X)$
is contractible.
$2)$ If $X$ is contractible or $Y$ then $\tilde{C}(X) \otimes
 \tilde{C}(Y)$ is contractible.
$3)$ If $X$ is contractible and $Y$ then $C(X) \otimes C(Y)$ is
acyclic.

If the first two descend from the first Lemma, i.e the first Lemma it's true with reduced complex, how the third part of the corollary could be proved ? My confusion here involves in particular involving augmented complex, since the definition of being contractible as chain complex is that all the homology (included $0$ degree) is trivial, which feels contradictory to me with what I'd like to proof and the definition of acyclic itself.
What's the standard definition of contractible chain complex ? What's the real statement of the Lemma and how the Corollary could be proven ?
Any help would be appreciated, I'd like to understand those topics because I'm quite confused.
 A: The definition of "contractible" is what's stated in the lemma: the identity map is chain homotopic to 0. This implies that the homology is 0 in all dimensions. It's a purely algebraic condition, so you can discuss whether any chain complex is contractible, but in particular it does apply to the augmented chain complex of a contractible space.
The lemma as stated is correct, and parts (1) and (2) of the corollary should follow immediately (as you say, applying the lemma to the augmented chain complexes, which will be contractible).
What does "acyclic" mean in part (3)? That the homology is $\mathbb{Z}$ in dimension 0, zero otherwise? It often means that all of the homology groups are zero. I don't know what was intended, but anyway, I would be tempted to prove (3) as follows (although I haven't checked any details): $\tilde{C}(X)$ being contractible says that $C(X) \to \mathbb{Z}$ is a chain homotopy equivalence (where $\mathbb{Z}$ is a chain complex with $\mathbb{Z}$ in dimension 0, zeroes elsewhere). Then $C(X) \otimes C' \to \mathbb{Z} \otimes C'$ will be a chain homotopy equivalence — this should be equivalent to the statement that $\tilde{C}(X) \otimes C'$ is contractible, i.e., the lemma. Then (applying the same thing twice) the composite
$$
C(X) \otimes C(Y) \to \mathbb{Z} \otimes C(Y) \to \mathbb{Z} \otimes \mathbb{Z}
$$
will be a chain homotopy equivalence.
I'm sure there are other approaches to (3), maybe deriving it directly from (2).
