Short exact sequence of pointed spaces induces isomorphism of quotients Let $X,Y,Z$ be pointed topological spaces. I will denote the point by $*$ and write $0=\{*\}$. Let $f:X\to Y$, $g:Y\to Z$ be continuous maps preserving the point. The sequence $X\xrightarrow{f} Y\xrightarrow{g} Z$ is called exact if im$(f)=g^{-1}(*)$. Thus, we can talk about short exact sequences. Let $0\to X\xrightarrow{f} Y\xrightarrow{g} Z\to 0$ be such a short exact sequence. Then $g$ is automatically surjective. Let's also assume that $f$ is injective.
Question 1: Is $Y/X$ homeomorphic to $Z$?
Question 2: Is there any reference for short exact sequences of pointed spaces? I'm especially interested in iterated extensions of $\mathbb{R}^n$.
My thoughts: I tried solving this categorically and I was able to prove that in the notation above, $f$ is an effective and normal monomorphism in the category of pointed topological spaces and $g$ is an effective and normal epimorphism. Thus, we have canonical factorizations $X\xrightarrow{\cong}\text{im}(f)\to Y$ of $f$ and $Y\to\text{coim}(g)\xrightarrow{\cong} Z$ of $g$, where $\text{im}(f)$ and $\text{coim}(g)$ denote the categorical image and coimage, respectively. (I'm not sure if $\text{im}(f)$ is the set-theoretic image in this case.) This means that $\text{im}(f)$ is the equalizer of the pushout morphisms $Y\to Y\sqcup_X Y\leftarrow Y$ and $\text{coim}(g)$ is the coequalizer of the pullback morphisms $Y\leftarrow Y\times_Z Y\to Y$. In particular, we can see that $X$ is equipped with the subspace topology of $Y$ and $Z$ is equipped with the quotient topology of $Y$, so this situation seems promising.
To show that $Y/X$ is isomorphic to $Z$, we have to show that $Z$ is the pushout of $f:X\to Y$ and the trivial map $u^X:X\to 0$. The pushout morphisms should be given by $g:Y\to Z$ and the trivial map $u_Z:0\to Z$, so by exactness, we already know that $g\circ f=u_Z\circ u^X$. So let $F$ be a pointed topological space and let $\psi_1:Y\to F$, $\psi_2:0\to F$ satisfy $\psi_1\circ f=\psi_2\circ u^X$. Obviously, $\psi_2$ is the trivial map $u_F$, so we just have to find a unique morphism $\rho:Z\to F$ such that $\psi_1=\rho\circ g$. Since $g$ is an epimorphism, we know that such a morphism is necessarily unique and we just need to show its existence. Since $Z$ is already a pushout, I think we need to use its universal property but I have no clue how to apply it in this situation. I believe the difficulty is to identify $X$ and $Y\times_Z Y$ in an appropriate way. Alternatively, since we know that $Z$ is some quotient of $Y$, one could maybe try to use the universal property of the quotient space to prove the existence of $\rho$. For this, one would have to show that $g(y)=g(y')$ implies $\psi_1(y)=\psi_1(y')$ but I don't see why this should be true. Any ideas?
 A: You should not write $Y/X$, but $Y/f(X)$. Only if you know that $f$ is an embedding, you could understand that $X$ is a genuine subspace of $Y$ and write $Y/X$. Anyway, you see that the topology of $X$ is irrelevant. Just consider
$$0  \to f(X) \hookrightarrow Y \stackrel{g}{\to} Z \to 0$$
where $f(X)$ carries the subspace topology inherited from $Y$.
Q1: In general no. Not even on the level of sets $g$ needs to induce a bijection $\bar g : Y/f(X) \to Z$. In fact, $g$ may have non-trivial fibers $g^{-1}(z)$ for all $z \in Z$.
But even if $\bar g$ is known to be a bijection, there is almost no relationship between the topologies of the spaces Consider for example $Z$ with the trivial topology. Then all functions $g : Y \to Z$ are continuous, but there is no reason to assume that $Y/f(X)$ has the trivial topology.
Q2: No. Short exact sequences of pointed spaces are not useful.
Update:
Let $X  = Z  = \mathbb R, Y = \mathbb R^2$, $f : X \to Y, f(x) = (x,0)$, and $g : Y \to Z$ be the projection onto the second coordinate. This is an open map, hence a quotient map. We have $Y / f(X)) \not\approx Z$.
