Let $M, N$ be smooth manifolds with appropriate dimensions, $X$ be a (smooth) vector-field on $M$.
Let $f: M \rightarrow N$ be a map. Now, what I know is that to pushforward $X$ on $N$, $f$ is required to be an injection; and that for 'pushed' $X$ on $N$ to be smooth, $f$ is required to be smooth. (given $X$ is a smooth vector field)
But I've seen people say that $f$ should be a diffeomorphism (i.e. $M\cong_{diff} f(M)$).
Does $f$ have to be a diffeomorphism? This naturally poses another question - Does a bijection that is smooth in one way, necessarily smooth the other way? (If yes, then $f$ is already a diffeomorphism)
Let $X\in\mathfrak{X}(M)$ be a smooth vector field on $M$. If all you have is a smooth bijection $F\colon M\to N$ with an inverse $F^{-1}$ that isn’t smooth, there is no way to guarantee that the resulting vector field $F_*X$ is smooth. That said, a smooth bijection is enough to define a unique rough (that is, not necessarily smooth) vector field.
Following the suggestion of @Tom Ariel in the comments, consider the (constant) coordinate vector field $X$ defined by $X_p=\partial/\partial x|_p$ on $\mathbf{R}$. Its pushforward under the map $F\colon\mathbf{R}\to\mathbf{R}$ defined by $F\colon x\mapsto x^3$ gives us a rough vector field $F_*X$ defined by $$\begin{align*} (F_*X)_p &=dF_{F^{-1}(p)}(X_{F^{-1}(p)}) \\ &=dF_{p^{1/3}}(\partial/\partial x|_{p^{1/3}}) \\ &=3p^{2/3}(\partial/\partial x|_p). \end{align*}$$ This vector field is not smooth; the map $x\mapsto 3x^{2/3}$ isn’t even differentiable at $x=0$.
However, if we have a diffeomorphism $G\colon M\to N$, then we are guaranteed that $G_*$ takes smooth vector fields to smooth vector fields, since $G_*X\colon N\to TN$ is then the composition of smooth maps $$N \overset{G^{-1}}{\longrightarrow} M \overset{X}{\longrightarrow} TM \overset{dG}{\longrightarrow} TN.$$
(For references, see for example Propositions 8.1 and 8.19 in John Lee’s Introduction to Smooth Manifolds.)