Example

Example 25

Consider the identity function \(I_{\mathbb{N}} : \mathbb{N} \rightarrow \mathbb{N}\) defined as \(I_{\mathbb{N}}(x) = x\), \(\forall\, x \in \mathbb{N}\).
Show that although \(I_{\mathbb{N}}\) is onto but \(I_{\mathbb{N}} + I_{\mathbb{N}} : \mathbb{N} \rightarrow \mathbb{N}\) defined as \[ (I_{\mathbb{N}} + I_{\mathbb{N}})(x) = I_{\mathbb{N}}(x) + I_{\mathbb{N}}(x) = x + x = 2x \] is not onto.