求证映射$f:X→Y$为嵌入映射当且仅当诱导同胚$ ilde f:X→f(X)$

Let's formalize and complete the proof you started, focusing on the equivalence between embeddings and homeomorphisms onto the image.

First, recall the key definitions:

• A map $f: X \to Y$ is an embedding if it is injective, continuous, and the topology on $X$ is exactly the initial topology with respect to $f$—meaning $\tau_X = {f^{-1}(V) \mid V \in \tau_Y}$.
• The induced map $\tilde{f}: X \to f(X)$ is just $f$ restricted to its image, where $f(X)$ carries the subspace topology inherited from $Y$.

Forward Direction: If $f$ is an embedding, then $\tilde{f}$ is a homeomorphism #

• Bijectivity: Since $f$ is injective, $\tilde{f}$ is a bijection between $X$ and $f(X)$.
• Continuity: For any open set $U$ in $f(X)$, by subspace topology, $U = V \cap f(X)$ for some $V \in \tau_Y$. Then $\tilde{f}^{-1}(U) = f^{-1}(V)$, which is open in $X$ by the embedding definition. So $\tilde{f}$ is continuous.
• Openness (hence inverse continuous): Take any open set $W \in \tau_X$. Since $f$ is an embedding, $W = f^{-1}(V)$ for some $V \in \tau_Y$. Then $\tilde{f}(W) = f(W) = f(f^{-1}(V)) = V \cap f(X)$, which is open in $f(X)$ (subspace topology). So $\tilde{f}$ is an open map, meaning its inverse $\tilde{f}^{-1}$ is continuous.

Combining these, $\tilde{f}$ is a homeomorphism.

Reverse Direction: If $\tilde{f}$ is a homeomorphism, then $f$ is an embedding #

We need to show $\tau_X = {f^{-1}(V) \mid V \in \tau_Y}$.

• One inclusion: Since $f$ is continuous (it's the composition of $\tilde{f}$ and the continuous inclusion map $i: f(X) \to Y$), every $f^{-1}(V)$ for $V \in \tau_Y$ is open in $X$. So ${f^{-1}(V) \mid V \in \tau_Y} \subseteq \tau_X$.
• Other inclusion: Take any $W \in \tau_X$. Since $\tilde{f}$ is a homeomorphism, $\tilde{f}(W)$ is open in $f(X)$, so there exists $V \in \tau_Y$ such that $\tilde{f}(W) = V \cap f(X)$. Then:
  $$f^{-1}(V) = \tilde{f}^{-1}(V \cap f(X)) = \tilde{f}^{-1}(\tilde{f}(W)) = W$$
  This means $W$ is in ${f^{-1}(V) \mid V \in \tau_Y}$, so $\tau_X \subseteq {f^{-1}(V) \mid V \in \tau_Y}$.

Thus the two topologies are equal, and since $\tilde{f}$ is bijective, $f$ is injective. Therefore $f$ is an embedding.

内容的提问来源于stack exchange，提问作者Francesco Gemma