I am trying to show that $\mathbb{D}^2$ doesn't retract to $S^1$. I am using the fact that if $X$ is a subspace of $Y$ then if there is a retract $r:X\rightarrow{Y}$ with $r\circ{i}=Id_{X}$, with $i$ being an inclusion, then $i_*:H_n(X)\rightarrow{H_n(Y)}$ is injective.
Since we have
(i) $H_n(S^1)=\mathbb{Z}$
(ii) $H_n(\mathbb{D}^2)=0$
clearly there is no such injection. But how do I show (i) and (ii)?
apologies if this is obvious.
note: im not looking for an alternative proof, just one that involves (i),(ii)
You have to have $n=1$ for (i) to hold, but of course that's enough.
Proving it sort of depends on what you already know.
For (i) :
If you know the Hurewicz theorem and that $\pi_1(S^1,1) = \mathbb Z$, then you are done
If you know about cellular/simplicial homology, then you can easily compute $H_1(S^1)$ via the celullar/simplicial complex associated to the following decomposition : one $0$-cell, and one $1$-cell.
If you know about the Mayer-Vietoris exact sequence, you can use it by choosing the opens to be two open intervals that cover $S^1$, e.g. $S^1$ minus the south pole and $S^1$ minus the north pole. Then you are reduced to knowing the homology of an interval. For this you can use any of the above $2$ methods, or the first method for $\mathbb D^2$ described below.
For (ii) :
If you know that homology is homotopy invariant, you can notice that $\mathbb D^2$ is contractible, i.e. homotopy equivalent to a point, hence they have the same homology, and then the homology of a point is easy to compute.
You can use any of the first two methods described for (i) above, if you know about them.