Let $\mathcal{C}$ be the category of toplogical(resp. differential) manifolds. The objects of $\mathcal{C}$ are topological(resp. differential) manifolds and the morphisms of $\mathcal{C}$ are continuous(resp. smooth) maps.
Let $f\colon X \rightarrow Y$ be a morphism in $\mathcal{C}$. We claim $f$ is a monomorphism if and only if $f$ is injective.
Suppose $f$ is a monomorphism. Let $x, y$ be distinct points of $X$. Let $p$ be a $0$-dimensional object in $\mathcal{C}$. There exists the unique morphism $g\colon p \rightarrow X$ such that $g(p) = x$. Similarly there exists the unique morphism $h\colon p \rightarrow X$ such that $h(p) = y$. Since $g \neq h$, $fg \neq fh$. Hence $f(x) \neq f(y)$. Hence $f$ is an injective map.
Conversly suppose $f$ is injective. Clearly $f$ is a monomorphism.
"Are they the same as submanifolds?"
Generally no.
Counter-example:
Let $f\colon \mathbb{R} \rightarrow \mathbb{R}^2$ be the map defined by $f(x) = (x^3, 0)$. $f$ is smooth and injective, but is not an immersion($f'(0) = 0$). Hence $\mathbb{R}$ cannot be identified with a submanifold of $\mathbb{R}^2$ by $f$.