Notation: isomorphism between summands When is a group homomorphism
$$\varphi:\Bbb Z/2\Bbb Z\to \Bbb Z/2\Bbb Z\oplus \Bbb Z/2\Bbb Z$$
an "isomorphism onto the first summand" ?
Is the map $\varphi:1\mapsto (1,1)$ an isomorphism onto the first summand?
 A: To say that a map
$$\varphi:A\to B\oplus C$$
is an isomorphism onto the first summand means that there is an isomorphism $\psi:A\to B$ for which $\varphi(a)=(\psi(a),0)$. 
In other words, $\varphi$ is an isomorphism onto the first summand when both of the following are true:


*

*the image of $\varphi$ is equal to the first summand, i.e. the subgroup (or subring, etc.) $B\oplus 0$ of $B\oplus C$, and moreover,

*considering the function $\varphi$ as a map solely onto its image (i.e., restricting the codomain of $\varphi$), it is an isomorphism.


Thus, the map $\varphi:\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ defined by $\varphi(1)=(1,1)$ is not an isomorphism onto the first summand, since the image of this $\varphi$ is not the subgroup $\mathbb{Z}/2\mathbb{Z}\oplus 0$ of $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.
A: Such a $\varphi$ is an isomorphism in case it is bijective onto its image and has its image contained in the first summand, which is the subset $\mathbb{Z}/2\mathbb{Z} \oplus 0 = \{ (0,0), (1,0) \}$. So your map is not such an isomorphism.
