$K$ is DG-algebra, $H_\star(K)$ forms graded $R-$algebra. Why $1\in H_0(K)$ or $[1]\neq 0\in H_0(K)$?

Question

This is related to Weibel Exercise 4.5.1

A graded $R-$algebra $K_\star$ is a positive complex(i.e. $\star\geq 0$) s.t. $K_p\otimes K_q\to K_{p+q}$ is $R-$bilinear and $1\in K_0$ s.t. $K_\star$ is associative algebra with unit. DG-algebra is graded $R-$algebra with differential behaving like differential on differential forms.

Let $K$ be a DG-algebra. Show that homology $H_\star(K)=\{H_p(K)\}$ forms a graded R-algebra.

$\textbf{Q:}$ It is easy to check associativity and $R-$product bilinearity due to cycle vanishing on differential. I have trouble to check $[1]\neq 0\in H_0(K)$. First note that $K_0\to 0$ implies $1\in H_0(K)$. However, I can't verify that $d(a)=1$ for some $a\in K_1$. If this is the case, then I have to conclude that $H_\star(K)$ is non-unital. Why $[1]\neq 0\in H_0(K)$?

Answer 1

It does not matter whether $[1]=0$. If $[1]=0$, the algebra is still unital, since $[1]$ still is a multiplicative identity.

Answer 2

Another way to see that if $d(a) = 1$ then the whole ring is $0$ is the following (although it is clear from Eric's answer) : assume $d(b) = 0$. Then $d(ab) = d(a)b = b$.

Therefore all cycles are boundaries : $H_*(K) = 0$.