Let $f:X \to Y$ be an injective map. Show that $f$ is an embedding iff $f$ induces the topology on $X$. 
Let $f:X \to Y$ be an injective map. Show that $f$ is an embedding iff $f$ induces the topology on $X$.

$”\implies”$ Assume that $f$ is an embedding. This means that $f$ is a homeomorphism onto it’s image. Now the topology on $f[X]$ is defined as $$\tau_{f[X]} = \{ f^{-1}[U] : U \in \tau_Y \} = \{ f[X] \cap U : U \in \tau_Y \}.$$
So the topology $\tau_X = \{f^{-1}[V] : V \in \tau_{f[X]} \}$, but $V \in \tau_{f[X]} $ is of the form $f[X] \cap U$, where $U \in \tau_Y$ so $$\tau_X = \{f[X] \cap U: U \in \tau_Y\}$$ also and thus induced by $f$?
$”\Longleftarrow”$ Assume that the topology $\tau_X$ is induced by the injection $f:X \to Y$. This means that $\tau_X = \{f[X] \cap U: U \in \tau_Y\}$ assuming $”\implies”$ was correct.
Now this is where I’m stuck. It seems that I need to use the  injectivity of $f$ somehow here? Any hints on how to proceed?
Edit: For $”\Longleftarrow”$ if the topology induced on $X$ is from $f: X \to f[X]$, then the conditions for $f$ to be an embedding are satisfied I think.
$(1)$ by definition $f$ is an injection.
$(2)$ the topology induced gives us that $f$ is continuous
$(3)$ for any $U$ open in $X$ by the induced topology $f[U]$ is open in $f[X]$.
So this would imply that $f$ is an embedding?
I’m not sure about the condition $(3)$ my book has it that $f^{-1} : f[X] \to X$ should be continuous, but it states immediately after that this can be replaced with the condition I proposed.
 A: If $f: X \to Y$ is an embedding then $f$ is a homeomorphism between $X$ and $f[X]$. In particular, $f: X \to f[X]$ is an open map. Now, if $O$ is open in $X$ we know that $f[O]$ is open in $f[X]$ so there is an open set $O’ \subseteq Y$ so that $f[O] = O’ \cap f[X]$.
I claim that $O = f^{-1}[O’]$: if $x \in O$, $f(x) \in O’$ so $x \in f^{—1}[O]$ while if $x \in f^{-1}[O’]$ and so $f(x) \in O’$ so also $f(x) \in f[O]$, as $f$ is 1-1 (!) we conclude that $x \in O$ and we’re done.
It follows that $O$ is in the topology on $X$ induced by $f$, being the inverse image of $O’ \in \tau_Y$.  This proves one direction.
Now suppose $X$ has the topology induced by $f$. Then $f: X \to f[X]$ is a bijection (1-1 being given, and onto by construction) and continuous (as $f$ is) so we only need that $f$ is open as a map from $X$ to $f[X]$. This is clear, as if $O \in \tau_X$ we have some $O’ \in \tau_Y$ so that $O = f^{-1}[O’]$ (definition of induced topology) and then we similarly see that $f[O] = O’ \cap f[X]$ and so $f[O]$ is open in $f[X]$, as required. So $f$ is a homeomorphism onto its image, hence an embedding.
