# Local injectivity on connected space

I have a solution to this problem but I would like to know your opinions (better, solutions). So I will not write the solution yet.

Here is the problem:

Let $M$ and $N$ be metric spaces, with $M$ connected. Suppose that $f:M\to N$ is a continuous and locally injective map and that there exists $g:N\to M$ a continuous map such that $f\circ g=Id:N\to N$.

Prove that $f$ is a homeomorphism.

Thanks for a while.

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This is a rather strange thing to ask for. I believe most of us will be of the opinion that $f$ is a homeomorphism. – Kevin Carlson Aug 19 '12 at 2:08
@KevinCarlson, sorry. Maybe I would ask for other solutions to compare with mine. Thanks for your attention. – Sigur Aug 19 '12 at 14:16

Step 1: For each $x\in N$, there is an open neighborhood $U_x$ of $x$ such that $V_x:=g(U_x)$ is open in $M$ and $g\colon U_x\to V_x$ is a homeomorphism.

Proof: Let $x\in N$, and let $V$ be a neighborhood of $g(x)$ in $M$ such that $f$ is injective on $V$. Let $U_x = g^{-1}(V)$; this is a neighborhood of $x$ in $N$. Let $V_x = f^{-1}(U_x)\cap V$; this is a neighborhood of $g(x)$ in $M$.

If $y\in U_x$, then by the definition of $U_x$ one has $g(y)\in V$. But also $f(g(y)) = y$, so $g(y)\in f^{-1}(U_x)$. This proves $g(y)\in f^{-1}(U_x)\cap V = V_x$, i.e., that $g$ maps $U_x$ into $V_x$. It also proves that $f$ maps $V_x$ surjectively onto $U_x$, since $y = f(g(y))\in f(V_x)$. Putting all of this together, we have derived that $f|_{V_x}\circ g|_{U_x} = Id|_{U_x}$, and, since $f|_{V_x}$ is injective, $g|_{U_x}\circ f|_{V_x} = Id|_{V_x}$.

Step 2: Since $f\circ g = Id$, we must have $g$ is injective. Combining this with step 1 yields that $g$ is a homeomorphism onto an open subset $g(N)$ of $M$.

Step 3: On the other hand, $g\circ f\colon M\to g(N)$ is a retraction of $M$ onto $g(N)$, and thus $g(N)$ is closed.

Step 4: Since $g(N)$ is both open and closed, $g(N) = M$, and hence by step 2, $g$ is a homeomorphism onto $g(N) = M$. The relation $f\circ g = Id$ shows that $f$ is the inverse of $g$, and hence $f$ is also a homeomorphism.

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Thanks. Your solution is essentially like mine. I considered the set $\{x\in M; g\circ f(x)=x\}$ and proved that it is non empty and both open and closed. – Sigur Aug 20 '12 at 0:21