Is $\varphi:G\rightarrow G $ ,$\varphi(a)=a^{-1}$ homomorphism? Let $G$ be a group and $\varphi:G\rightarrow G $    ,$\varphi(a)=a^{-1}$
I need to find if $\varphi$ is homomorphism, and if so to find $ker(\varphi)$ and to tell if $\varphi$ is one-to-one and\or onto.
My Attemept:
$\varphi(aa^{-1})=\varphi(e)$ because $aa^{-1}=e$
$\varphi(a)\varphi(a^{-1})=a^{-1}a$ by defenition $=e$
so it is homomorphism (I think so)
 A: Is $\varphi$ a homomorphism?: $\varphi(ab) = (ab)^{-1} = b^{-1}a^{-1} = \varphi(b)\varphi(a)$. Here you see that the image of $\varphi$ needs to be Abelian for $\varphi$ to be a homomorphism. So what is the image of $\varphi$? Well, since every element in $G$ has an inverse, then $\varphi(G) = G$.
In all you need $G$ to be Abelian for $\varphi$ to be a homomorphism. As @Tobias mentions above, this is actually an if and only if.
A: the mapping $\phi$ is an isomorphism of $G$ to its opposite group $G^{op}$ (think mirror images). if $G$ is abelian the distinction between $G$ and $G^{op}$ collapses since:
$$
a \circ^{op} b = b \circ a = a \circ b
$$
A: Let us check homomorphism first what is the definition of homomorphism?
assuming group operation is multiplication we have.
$\varphi(ab) = \varphi(a)*\varphi(b)$ That is the definition of homomorphism.
Let us check if it does indeed satisfy it:
$\varphi(a)*\varphi(b) = a^{-1} * b^{-1}$ and $\varphi(ab) = (ab)^{-1} = b^{-1}a^{-1}$ 
So it is homomorphism iff $G$ is abelian you can easily prove it.
Now Let us assume that G is abelian and suppose that we have the following
$\varphi(a) = \varphi(b) \rightarrow a^{-1} = b^{-1}$
Hence inverting both side we get $a = b$, so it is injective.
A: If $φ$ is homomorphism then for every $a,b$ belong to G $ab=φ((ab)^{-1})=φ(b^{-1}a^{-1})=φ(b^{-1})φ(a^{-1})=ba$ so G is abelian.  
