Prove some properties of continuous functions whose composition is an identity map Let X and Y be topological spaces, and let $f:X \rightarrow Y$, $g:Y \rightarrow X$ be continuous maps such that $fog$ is the identity map onto itself.  Prove the following statements:
(a) $f$ is onto and g is one-to-one
(b) Y has the quotient topology determined by $f$
(c) $g$ maps Y homeomorphically onto a subspace of X
(d) If X is a Hausdorff space, then so is Y
My attempts
(a) Let $y \in Y$, choose $x=g(y)$, then $f(x)=f(g(y))=(fog)(y)=y$ so f is surjective.
Suppose $g(a)=g(b)$, then $f(g(a))=f(g(b))$ so $a=b$, therefore $g$ is injective
(b) We show that f is an open map.  Let U be open in X.  Since g is continuous, $g^{-1}[U]$ us open in Y.  Now $fog$ is the identity map so we have the following
$g^{-1}[U]=id_Y(g^{-1}[U])=(fog)(g^{-1}[U])=f[U]$, where the latter holds since g is surjective.  Therefore $f[U]$ is open in Y so f is an open map.  Hence f is a quotient map and it follows that Y has the quotient topology determined by $f$.
(c)  We first show that $g$ is an open map.  Let U be open in Y, then $f^{-1}[U]$ is open in X.  Now $f^{-1}[U]=(f^{-1}oid_Y)[U]=(f^{-1}ofog)[U]=g[U]$, so g is an open map.
We have that g is a continuous, open and injective map.  It follows that Y is homeomorphic to the image of g which is a subspace of X with the subspace topology.
d) Suppose X is Hausdorff.  Let $y_1$ and $y_2$ be two distinct points in Y.  Since g is injective, $g(y_1)$ and $g(y_2)$ are distinct points in X.  Consequently there exists disjoint open sets $G_1$ and $G_2$ containing $g(y_1)$ and $g(y_2)$ respectively.  Since g is continuous, $g^{-1}[G_1]$ and $g^{-1}[G_2]$ are open neighborhoods of $y_1$ and $y_2$ respectively.  Furthermore g is injective thus $g^{-1}[G_1]\cap g^{-1}[G_2]=\emptyset$.  Hence Y is Hausdorff.
Is this correct?
Thank you

Proof of Part (c) is incorrect. Because you used $f^{-1}o f=Id_X$, but this is in general not true because $f$ need not to be injective.
The map $g:Y\to X$ need not to be open map. Think about a trivial example $f:\mathbb R^2\to \mathbb R$ by projection onto the first coordinate $(x,y)\mapsto x$, and $g:\mathbb R\to \mathbb R^2$ by inclusion  $x\mapsto (x,0)$. Clearly, the image of open set in $\mathbb R$ under map $g$ is not open.
Nevertheless, $g$ still maps $Y$ homomorphically onto its image $g(Y)$, where $g(Y)$ induces subspace topology from $X$. To see this, notice $g(U)=f^{-1}(U)\cap g(Y)$, therefore $g(U)$ is open in $g(Y)$. Thus $g$ is open map onto its image, and also $g$ is one-to-one, then $g$ is homeomorphism onto its image.
