A Homology calculation Question

Question

Let $q: S^n\rightarrow S^n\vee S^n$ be the map we get quotienting the equator. What is $q_*$ on $n$-th Homology level? $q_*:\mathbf{Z}\rightarrow \mathbf{Z}\oplus\mathbf{Z}$ ?

Answer 1

If you follow up by collapsing say the second $S^n$ you get a map $S^n\to S^n$ which consists in collapsing the top hemisphere to a single point.

However the top hemisphere is a contractible sub-CW-complex, hence collapsing it is a homotopy equivalence. Hence the induced map in homology is an isomorphism $\mathbb{Z\to Z}$ ($1$ or $-1$ depending on orientation but you can't really make a difference at that level).

Of course it's the same thing if you follow up by collapsing the first $S^n$.

Now use the fact that $H_q(X\vee Y)\to H_q(X)\oplus H_q(Y)$ obtained by summing the two "collapsing" maps is an isomorphism to conclude that your map is simply $\mathbb{Z\to Z\oplus Z}, x\mapsto (x,x)$ (with possible minus signs, but again, at this level you can't really make a difference)

Answer 2

In addition to Max's answer I think this is how we can figure out the signs properly if we fix the convention as mentioned in the comments of Max's answer.

As pointed out by Max, to determine the map to each $\mathbf{Z}$ factor, we quotient out the other hemisphere and treat it as a map from $S^n$ to $S^n$. Now quotienting out the lower hemisphere will give us the map into the first factor of $\mathbf{Z}\oplus\mathbf{Z}$, which is $\mathbf{Z}\rightarrow\mathbf{Z}$; $x\mapsto x$. Now quotienting out the upper hemisphere will actually give us the map $x\mapsto -x$, because we can think of this map as first applying a rotation to $S^n$ interchanging the positions of the hemispheres, which takes $\Delta_1-\Delta_2$ to $\Delta_2-\Delta_1$. So this gives a $-1$ degree map, and then quotienting out the lower hemisphere gives identity from $\mathbf{Z}\rightarrow \mathbf{Z}$, so the composition is $x\mapsto-x$. So, $q_*$ should take $x$ to $(x,-x)$