Why must $Y$ be mapped homemorphically into some adjunction space $X\cup_fY$ Let $X$ be a topological space with a subspace $A$. Let's say we have a continuous map $f:A\rightarrow Y$. The adjunction space $X\cup_fY$ can be taken to be a quotient of $X\coprod Y$. Let $p:X\coprod Y\rightarrow X\cup_fY$ be the quotient map, let $t_1:X\rightarrow X\coprod Y,t_2:Y\rightarrow X\coprod Y$ be the natural embeddings. My book says that $pt_2$ must map $Y$ homemorphically to its image in $X\cup_f Y$. I don't see why this has to be true. So far I can only see that $pt_2$ is injective. I ll appreciate if someone can give me an argument
Thank you  
 A: Surjectivity of $pt_2$ follows since we are saying that $pt_2$ is homeomorphic onto its image. Thus a set level inverse exists to $pt_2$ when restricted to the image, call it $g: B \to Y$, where $B = pt_2(Y)$.
We now need only show $g$ is continuous. I may be able to come back and write this up fully later, but the most common thing people overlook is that we need to show $g$ is continuous in the subspace topology. This means that if we are given an open set $U \subset Y$, we need to show $g^{-1}(U) = pt_2(U)$ is open in the subspace topology, meaning that $g^{-1}(U) = pt_2(U) = B \cap V$ for some open $V \subset X \cup_f Y$.
For ease of notation, let's define $X \coprod Y = X \times \{0\} \cup Y \times \{1\}$.  
Let $U \subset Y$. Then $U \times \{1\} = t_2(U)$ is an open set in $X \coprod Y$. Now, the quotient topology on $X \cup_f Y$ is defined so that $p: X \coprod Y \to X \cup_f Y$ is continuous. Then since $p(U \times \{1\}) = p(Y) \cap S$, where $S = \{[(a, 0)] : f(a) = y$ for some $y \in U\}\cup \{[(u, 1)] \in U \times \{1\} \}$, where brackets [] are taken as equivalence classes, pt_2(U) is open in the image, where S is open in $X \cup_f Y$ since $p^{-1}(S) = i^{-1}(U) \times \{0\} \cup U \times \{1\}$. 
