The following diagram, to call this set as $A$, is constructed from connecting topologist's sine curve and a set $\{(0,y)|-1 \le y \le 1\}$ by some path.
Intuitively $H_2(A) =0$.
But I don't know how to reason it since it is not a CW-complex.
How can I prove this?
Thanks.
Let $B$ be the subspace consisting of the T intersection and some of the sine curve next to it. Then we have the exact sequence
$$H_2(B)\rightarrow H_2(A)\rightarrow H_2(A,B)$$
Now $H_2(A,B)=0$ because we can excise the "bad" part and just get a line segment. On the other hand $B$ is connected but not pathwise connected, so the image of a singular simplex is either in the T intersection part or in the sine curve part. It follows that $H_2(B)=0$. Thus $H_2(A)=0$.