an injective map is embedding iff its left composition with any continuous map is continuous

Question

Let $X,Y$ be topological spaces and $i: X\rightarrow Y $ be injective map.
Then $i$ is an embedding $\Leftrightarrow$ for any set maps $f:Z\rightarrow X$, $f$ is continuous if and only if $i\circ f$ is continuous.
I am not sure if what I am thinking is correct. To show the above result, suppose the RHS to be true and we need to see $i$ is an embedding. That is both $i$ and its inverse are continuous. To see it has a continuous inverse, we need to see $i(U)$ is open if $U$ is open. Is it correct? And what is the correct way to prove this result?

Answer 1

Let us first introduce precise notation.

Let $A, B$ be topological spaces. We use the word function for any set-theoretic function $\phi : A \to B$ between the underlying sets. If $\phi$ is continuous, we explicitly say that $\phi$ is a continuous function.

For each function $\phi : A \to B$ we denote by $\phi' : A \stackrel{\phi}{\to} \phi(A)$ the function obtained by restricting the codomain. Here $\phi(A)$ carries the subspace topology inherited from $B$. Note that $\phi'$ is surjective and if $\phi$ is continuous, then also $\phi'$ is continuous. Therefore, if $\phi$ is a continuous injection (= injective function), then $\phi'$ is a continuous bijection (= bijective function).

An emdedding is a continuous injection $\phi : A \to B$ such that the function $\phi' : A \to \phi(A)$ is a homeomorphism. The latter is equivalent to the requirement that $(\phi')^{-1} : \phi(A) \to A$ is continuous .

Now let $i : X \to Y$ be an injective function between topological spaces $X,Y$, Let us prove

• $i$ is an embedding $\iff$ For each function $f:Z\rightarrow A$ defined on a topological space $Z$, $f$ is continuous if and only if $i\circ f$ is continuous.

"$\implies$":

Let $f$ be continuous. Since $i$ is continuous (recall the definiton of "embedding"), we see that $i\circ f$ is continuous.

Let $i \circ f$ be continuous. We want to show that $f$ is continuous, i.e that $f^{-1}(U)$ is open in $Z$ for all open $U \subset X$. Since $i$ is an embedding, $i(U)$ is an open subset of the subspace $X' =i(X) \subset Y$. Write $i(U) = X' \cap V$ with an open $V \subset Y$. We know that $(i \circ f)^{-1}(V)$ is open in $Z$. But $(i \circ f)^{-1}(V) = f^{-1}(i^{-1}(V)) = f^{-1}(U)$.

"$\Longleftarrow$":

1. $i$ is continuous: Take $Z = X$ and $f =id_X$. This is a continuous function, thus also $i = i \circ f = i \circ id_X$ is continuous.

2. $(i')^{-1}$ is continuous (which implies that $i$ is an embedding): Take $Z = i(X)$ and $f = (i')^{-1}$. The function $i \circ f = i \circ (i')^{-1}: i(X) \to Y$ is the subspace inclusion of $i(X)$ into $Y$ which is continuous. We conclude that $(i')^{-1} = f$ is continuous.