Prove that any map $T → T$ ($T$ = Torus) whose restriction to $S^1 ∨ S^1$ is null-homotopic induces a $0$ map on reduced homology

Question

Prove that any map $T → T$ ($T$ = Torus) whose restriction to $S^1 ∨ S^1$ is null-homotopic induces a $0$ map on reduced homology.

A few informations which I know :

(1) $T$ is obtained by attaching a 2-cell to it's 1-skeleton $S^1 ∨ S^1$ and thus $T / S^1 ∨ S^1 \simeq S^2$ .

(2) The quotient map $S^1 \times S^1 \to S^2$ collapsing the subspace $S^1 ∨ S^1$ to a point is not null homotopic and it induces an isomorphism on $H_2$ .

(3) Any map $S^2 \to S^1 \times S^1$ is null homotopic.

(4) Let $X$ be a simply connected space. Aany map from $X \to {(S^1 )}^k$ induces a 0 map on reduced homology.

Can I use these to come up with something? I have no idea! I get really confused when the problem asks to find the effect of some map on the homology.

Thanks in advance for help!

Answer 1

If $f:T\to T$ is nullhomotopic when restricted to $S^1\vee S^1$ then since the inclusion $S^1\vee S^1\to T$ is a cofibration (the inclusion of a skeleton in a CW-complex), you can extend that homotopy to $T$ and thus in fact assume that $f$ is constant on $S^1\vee S^1$.

It follows that it factors through $T/(S^1\vee S^1)\simeq S^2$.

Now, as Tyrone suggests, we may look at the universal cover of $S^1\times S^1$, which is just $\mathbb R^2$, hence a contractible space : its $\pi_2$ is trivial and therefore so is $\pi_2(S^1\times S^1)$ : in particular a map $S^2\to S^1\times S^1$ is nullhomotopic, therefore must induce $0$ on homology : so must the original map.

Answer 2

It is sufficient to look at the 1st and 2nd homology groups.

For the first you have the sequence $$S^1\lor S^1\rightarrow T \rightarrow T$$ This composition is null-homotopic . So at the level of homology we have

$$ \require{AMScd} \begin{CD} H_1(S^1\lor S^1)@>{i_*}>> H_1(T) @>{f_*}>> H_1(T) \end{CD} $$ the composition is $0$.

However we know $ \require{AMScd} \begin{CD} H_1(S^1\lor S^1)@>{i_*}>> H_1(T) \end{CD} $ is an isomorphism (you can see this from cellular homology ) and hence $$ \require{AMScd} \begin{CD} \ H_1(T) @>{f_*}>> H_1(T) \end{CD} $$ is the $0$ map. However you will run into problem at the $2$nd homology group if you try via this kind of argument.

So we approach as follows.

Now $(T, S^1\lor S^1)$ satisfy the Homotopy Extension Property. So if you have a map $$f: T\rightarrow T$$ and a homotopy $$H : S^1\lor S^1 \times [0,1] \rightarrow T$$ between $f|_{S^1\lor S^1}$ and $constant \ map$ you can extend this to get a homotopy between $f$ and $g$ where $g|_{S^1\lor S^1}$ is constant. Moreover $f_*=g_*$ by homotopy invariance property of homology.

Now by universal property of quotient topology $g$ factors through $$ \require{AMScd} \begin{CD} T@>{q}>>T/S^1\lor S^1\cong S^2 @>{\bar g}>> T \end{CD} $$

Now look at $\bar g:S^2\rightarrow T$. By general lifting lemma you can lift this to get a commutative diagram

$$\require{AMScd} \begin{CD} S^2@>{\tilde {\bar g}}>> \mathbb R\times \mathbb R\\ @VV{id}V @VV{p \ \ \ \ \ \ covering \ \ \ map}V \\ S^2 @>{\bar g}>> T \end{CD} $$

Since $\mathbb R^2$ is contractible $\bar g$ is nullhomotopic and hence $\bar g_*=0$

So $g_*=\bar g_*q_*=0$ As such $f_*=0$