I have been thinking about this question :
Show that if $i: N\rightarrow M$ is an invertible immersion then it is a diffeomorphism. Give a counterexample when $N$ is not Second Countable.
First time I was thinking about this my idea was to take the local coordinates for each $p$ such that we have $\psi\circ i\circ \phi^{-1}(x_1,...,x_n)=(x_1,...,x_n,0,...,0).$ Now without thinking much about it I thought that this would have to give me that $\dim M=\dim N$ and so in local coordinates we would have the identity and so we could prove that $i^{-1}$ is a smooth map . Now since this nowhere used the fact that $N$ is second countable I came up with a proof that does use this fact , using the fact that if $\dim N < \dim M$ then $i(N)$ has measure zero in $M$, which would contradict the fact that $i(N)=M$. Now here we use the fact that $N$ is second countable. But I can't find a reason why my first argument wouldn't work, so any help with that is appreciated. Thanks in advance.
The problem is that from the local structure of $i$ you cannot deduce that $\dim M = \dim N$. You know that locally around a point $p \in N$ and $i(p) \in M$, the map looks like $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_n,0,\dots,0)$ so the point in $M$ corresponding to $(0,\dots,0,\varepsilon,\dots,\varepsilon)$ for some $\varepsilon > 0$ definitely doesn't have a preimage under $i$ in a neighborhood of $p$ but a priori it might have a preimage far away from $p$.
To see that this is indeed possible, take $N$ to be $\mathbb{R}$ with the discrete topology, $M$ to be $\mathbb{R}$ with the usual topology and $i$ the (set theoretic) identity map. The manifold $N$ is not second countable, zero dimensional, the map $i$ is smooth, invertible (as a set map) and an immersion (as for any $p \in N$, the map $di|_{p} \colon T_pN \rightarrow T_pM$ is the only linear map from a zero dimensional space to a one-dimensional space which is injective). However, $N$ and $M$ are not diffeomorphic.