1)If $f:X→Y$ is injective then $f$ has left inverse. => Define $g:Y→X$ s.t. $g(y)= x$ when $y=f(x)$ and $X\circ$ otherwise Clearly $g\circ f =Ix$ and we are done.
The doubt is to imply this in example. What I have understood so far is I m gonna explain in this example. Consider $X={1, 2, 3}$ and $Y ={1,2,3,4}$ Given $f: X→Y$ s.t. $f(x)= x+1$ is an injective map. We need to find the left inverse of $f$ For that define $g:Y→X$ s.t. $g(y)= x$ when $y=x+1$ and $1$ otherwise Then $g$ is our left inverse here as $gof=Ix$
But we can define our 'otherwise' elements as $2$ or $3$ which will also be inverse of $f$ which is contradiction to the fact that if f has left inverse then it must be Unique.
I know I m misunderstanding something but not able to know actually what is wrong here ... majorly the problem is to map 'otherwise' terms.