If $f \circ g$ is onto then $f$ is onto and if $f \circ g$ is one-to-one then $g$ is one-to-one
This post intends to remove this question from the Unanswered list.
As noted in the comments, all of your assertions are correct.
This post intends to remove this question from the Unanswered list.
As noted in the comments, all of your assertions are correct.