How do I show that two maps are not homotopic. More precisely, how do I show that a map is not null-homotopic.

1-If $f,g:S^n \to S^n$ then its degree should be same. if their degree isn't same then they aren't homotopy equivalent.

2-Since homotopic maps induce same maps on fundamental groups, then if $f$ is null-homotopic, then induces the map that a constant map does. So If I am able to show that it doesn't induce the trivial map then it cannot be null-homotopic.

Can I see an example of a map which doesn't induce trivial map on the level of fundamental groups?

I am thinking about showing that the map $id: S^1 \to S^1$ is not null-homotopic. it seems to me that on the level of fundamental groups the map $$\gamma:[0,1] \to S^1$$ is sent to the generator $$[id \circ \gamma:[0,1] \to S^1\to S^1$$ is again the generator. Hence it sends $1 \to 1$ on the other hand the constant map sends $1$ to $0$ on level of fundamental groups. Is it good enough to conclude that $S^1$ is not contractible. I am wondering about examples of other spaces where I can show it is not null-homotopic.

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    $\begingroup$ Your observations about the the identity map $S^1 \to S^1$ are correct, from which you can conclude that it is not nullhomotopic. More generally, if $\pi_1(X) \neq 0$, then $\operatorname{id}_X$ is not nullhomotopic for the same reason. $\endgroup$ Commented Dec 25, 2022 at 13:55
  • $\begingroup$ You have already given an example of a map which doesn't induce a trivial map on the fundamental group, namly Id. We know Id is not null homotopic because S1 is not contractible, as evidenced by its fundamental group. Since you have already asked this question, or one quite like it, elsewhere on this site, is there something we do to help answer this is a way that hasn't already been tried? $\endgroup$ Commented Dec 25, 2022 at 13:56
  • $\begingroup$ In general, a space is contractible iff the identity is null-homotopic. So if the fundamental group is nontrivial, the space is not contractible, and so the identity map is not null-homotopic. $\endgroup$ Commented Dec 25, 2022 at 14:00
  • $\begingroup$ @IsAdisplayName, can you furnish some more interesting examples? This one was quite trivial and uninteresting $\endgroup$ Commented Dec 25, 2022 at 14:00
  • $\begingroup$ @IsAdisplayName, its not about the identity map only. I am interested in maps which are not identity as well $\endgroup$ Commented Dec 25, 2022 at 14:01

2 Answers 2


In this post you have two related, though distinct, questions: (i) when is a map between spaces null-homotopic and (ii) when does a map induce (or not induce) the trivial morphism between fundamental groups.

First I will answer (i). As you have already noted, $Id_{S^1}$ induces a non-trivial morphism of fundamental groups. There is a more general concept at play here:

Claim: for a space $X$, the identity on $X$ is null-homotopic if, and only if, $X$ is homotopy equivalent to the point.

The proof is rather trivial and is a good exercise. This claim characterizes when the identity in null homotopic. So we can take any non-contractible space, like $S^n$, and its identity is not null-homotopic. We'll come back to this example later, specifically when we want to see that null-homotpic maps and maps that induce trivial morphisms between fundamental groups are not the same.

But it seems like you want a more interesting example of a map that is not null-homotopic. Lets consider the bouquet of two circles, call it $A$ for brevity. This space can be constructed as wedge sum of $S^1$ with itself. As such, there are two canonical maps $inl,inr : S^1 \to A$. The first map picks out the loop "on the left" while the other picks out the "loop on the right" (see the link for a full description of the space). I claim that neither of these maps is null-homotopic. I will show this for $inl$, but the same argument applies to $inr$.

Claim: $inl$ is not null-homotopic.

Proof Outline: We can use the SvK theorem to calculate the fundamental group of $A$, which turns out to be the free group on two generators $\mathbb{Z}\star\mathbb{Z}$. We assume that $inl$ is null-homotopic, and then argue that $A$ must have the fundamental group of $S^1$. In fact, by assuming $inl$ is null-homotopic, you should be able to show that $inr$ is a homotopy equivalence. The idea here is that if $inl$ is null-homotopic, then the left hand-loop is contracted away and we are only left with the right-hand loop. So $inr$ should be a homotopy equivalence. Writing out the detail of this would be tedious, so I won't, but you can construct the necessary homotopy inverse and homotopies using the universal properties. Now, we know the fundamental group of $S^1$ is the free group on one generator $\mathbb{Z}$. Group theory tells us this is not isomorphic to the fundamental group on two generators. So we have arrived at a contradiction.

This proof outline hints at a general method for showing that a map is not null-homotopic: assume it is null-homotopic and then argue that this contradicts previously established results (the answer by user diracdeltafunk list a hefty slew of such calculations you can use for this purpose). This further hints at a reason why a map is (or is not) null-homotopic: a map is null-homotopic if it destroys all interesting paths and higher paths, and is not null-homotopic if it preserves some interesting path/higher path. I hope this helps answer (i).

Now on to (ii). Sticking with the above example, we can show $inl,inr:S^1\to A$ both induce non-trivial maps between fundamental groups. We have that $\pi_1(inr): \mathbb{Z}\to \mathbb{Z}\star \mathbb{Z}$ is the morphism given by the definition of $\mathbb{Z}\star\mathbb{Z}$ as the coproduct in $Grp$. That is, $\pi_1(inr)$ includes $\mathbb{Z}$ as "the left" copy of $\mathbb{Z}$ in $\mathbb{Z}\star\mathbb{Z}$. We can similarly cook up other examples, though it might be clear enough how to that I leave the other examples to you.

That is all well and good, but I have yet to come back to my claim claim that trivial maps between fundamental groups and null-homotopic maps are distinct concepts. For this, consider $Id:S^2 \to S^2$. By the first claim, we know that $Id$ is not null-homotopic. But, since $\pi_1(S^2)=0$, $Id$ necessarily induces the trivial morphism between fundamental groups. This is a not very interesting example of such a map.

We can cook up a more interesting example. Consider $S^1\vee S^2$, the wedge sum of $S^1$ with $S^2$. We can define a map $S^1\vee S^2 \to S^1\vee S^2$ that is a constant map on $S^1$ and is the identity on $S^2$ using the universal property of the wedge sum. Using the SvK theorem, $S^1\vee S^2$ has fundamental group $\mathbb{Z}$. But we the map we defined will be the trivial morphism. However, it will not be null homotopic. This helps illustrate a point I brought up earlier. The map we defined in not-null homotopic essentially because it preserves the non-trivial two dimensional loop of $S^2$. However, the map destroys all interesting 1-dimensional loops, which is why it induces a trivial morphism of fundamental groups.

I hope this helps!

  • $\begingroup$ when you said "assuming $inl$ is null-homotopic, you should be able to show that $inr$..." . Are you really sure it's correct or did you mean "assuming $inl$ is null-homotopic, you should be able to show that $inl$ is homotopy equivalence"? $\endgroup$ Commented Dec 25, 2022 at 16:30
  • $\begingroup$ @permutation_matrix Yes, the original is correct. Maybe I should write out some of the details of what I meant? The basic idea is that if $inl$ is null-homotopic, we can forget about the left-hand loop and only deal with the right-hand. Then, since we only have one loop, this will be just $S^1$. Also note that a null-homotopic map is homotopy equivalence iff the spaces in question are contractible. $\endgroup$ Commented Dec 25, 2022 at 16:45
  • $\begingroup$ @permutation_matrix I made an update to that section of the post. Lmk if that's not clear still, or if I should add more detail $\endgroup$ Commented Dec 25, 2022 at 18:12
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    $\begingroup$ I will be able to do now. If not I will surely let you know $\endgroup$ Commented Dec 25, 2022 at 18:19

If I am interpreting your question correctly, you are looking for examples of homotopy invariants of maps. Here are some.

  1. Action on homotopy groups. Any map $f : X \to Y$ of pointed spaces induces a map $\pi_n(f) : \pi_n(X) \to \pi_n(Y)$ for all $n \geq 1$. If $f \sim g$, then $\pi_n(f) = \pi_n(g)$ for all $n $. So, if $\pi_n(f) \neq \pi_n(g)$ for some $n$, then $f \nsim g$. This generalizes your observation about fundamental groups.
  2. Action on homology. Any map $f : X \to Y$ of spaces induces a map $H_n(f;A) : H_n(X;A) \to H_n(Y;A)$ for all $n \geq 0$ and all abelian groups $A$. If $f \sim g$, then $H_n(f;A) = H_n(g;A)$ for all $n$ and all abelian groups $A$. So, if $H_n(f;A) \neq H_n(g;A)$ for some $n$ and some abelian group $A$, then $f \nsim g$.
  3. Action on cohomology. Any map $f : X \to Y$ of spaces induces a map $H^n(f;A) : H^n(Y;A) \to H^n(X;A)$ for all $n \geq 0$ and all abelian groups $A$. If $f \sim g$, then $H^n(f;A) = H^n(g;A)$ for all $n$ and all abelian groups $A$. So, if $H^n(f;A) \neq H^n(g;A)$ for some $n$ and some abelian group $A$, then $f \nsim g$.
  4. Lefschetz number. Let $X$ be a space such that $\sum_{n \geq 0} \dim_{\mathbb{Q}} H_n(X;\mathbb{Q}) < \infty$. Then for any map $f : X \to X$, we can define $L(f) = \sum_{n \geq 0} (-1)^n \operatorname{Tr}_{\mathbb{Q}} H_n(f;\mathbb{Q})$. If $f \sim g$, then $H_n(f;\mathbb{Q}) = H_n(g;\mathbb{Q})$ for all $n$, so in particular $L(f) = L(g)$. Thus, if $L(f) \neq L(g)$, then $f \nsim g$. The Lefschetz number has two other interesting properties: it is always an integer, and if $X$ is compact and triangulable then $L(f) \neq 0$ implies that $f$ has at least one fixed point.

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