showing a map is not null-homotopic 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.
 A: 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!
A: If I am interpreting your question correctly, you are looking for examples of homotopy invariants of maps. Here are some.

*

*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.

*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 $f \nsim g$.

*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 $f \nsim g$.

*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.

