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**Drexel28** I came up with the following question, and would appreciate if I could get some validation/invalidation.

If $\displaystyle X$ is connected Hausdorff and $\displaystyle \varphi:X\mapsto Y$ continuous then is $\displaystyle \Gamma_\varphi\subseteq X\times Y$ connected? ($\displaystyle \Gamma_\varphi$ is the graph of $\displaystyle \varphi$). I think the answer is yes.

**Proof:** We first need a lemma

**Lemma:** Let $\displaystyle X$ be connected Hausdorff, then $\displaystyle \Delta_X\subseteq X\times X$ is connected ($\displaystyle \Delta_X$ is the diagonal).

**Proof:** Clearly the map $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ given by $\displaystyle x\mapsto (x,x)$ is continuous. But, it is not hard to see that $\displaystyle \Delta_X=\left(\iota_X\oplus \iota_X\right)\left(X\right)$. But, connectedness is invariant under continuous mapping. So, the conclusion follows $\displaystyle \blacksquare$

Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since $\displaystyle \varphi$ and the identity map $\displaystyle \iota_X$ are continuous we have that $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ is continuous. I now claim that $\displaystyle \Gamma_\varphi=\left(\iota_X\times\varphi\right)\l eft(\Delta_X\right)$.

To see this, let $\displaystyle (y,y')\in\Gamma_\varphi$ then $\displaystyle (y,y')=(x,\varphi(x))$ for some $\displaystyle x\in X$. Clearly then $\displaystyle \left(\iota_X\times\varphi\right)\left((x,x)\right )=\left(x,\varphi(x)\right)=(y,y')$ and so $\displaystyle (y,y')\in \left(\iota_X\times\varphi\right)\left(\Delta_x\ri ght)$

Conversely, let $\displaystyle (y,y')\in\left(\varphi\times\iota_X\right)\left(\D elta_X\right)$. Then, $\displaystyle (y,y')=\left(\varphi\times \iota_X\right)\left((x,x)\right)$ for some $\displaystyle x\in X$ but that means that $\displaystyle (y,y')=(x,\varphi(x))$ and so $\displaystyle (y,y')\in\Gamma_{\varphi}$.

Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. $\displaystyle \blacksquare$

Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.